Learning Objectives

  • Solve integration problems involving products of powers of \ds \text{sin}(x) and \ds \text{cos}(x).
  • Integrate products of sines and cosines of different angles.
  • Solve integration problems involving products of powers of \ds \text{tan}(x) and \ds \text{sec}(x).
  • Use reduction formulas to evaluate trigonometric integrals.

In this section we look at how to integrate a variety of products of trigonometric functions. These integrals are called trigonometric integrals. They are an important part of the integration technique called trigonometric substitution used for integrating functions involving certain root expressions that will be discussed in the next section. In addition, these types of integrals appear frequently when we study polar, cylindrical, and spherical coordinate systems later. Let’s begin our study with products of powers of \ds \text{sin}\phantom{\rule{0.1em}{0ex}}(x) and \ds \text{cos}\phantom{\rule{0.1em}{0ex}}(x).

Integrating Products and Powers of sin(x) and cos(x)

A key idea behind the strategy used to integrate combinations of powers of \ds \text{sin}\phantom{\rule{0.1em}{0ex}}(x) and \ds \text{cos}\phantom{\rule{0.1em}{0ex}}(x) involves rewriting these expressions as sums and differences of integrals of the form \ds {\int }^{}{\text{sin}}^{j}(x)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx or \ds {\int }^{}{\text{cos}}^{j}(x)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx that can be evaluated using u-substitution. Before describing the general process in detail, let’s take a look at the following examples.

Evaluating \ds  {\int }^{}{\text{cos}}^{j}(x)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Evaluate \ds {\int }^{}{\text{cos}}^{3}(x)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Solution

Make a substitution \ds u=\text{cos}\phantom{\rule{0.1em}{0ex}}(x). In this case, \ds du= - \text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx . Thus,

\ds \begin{array}{cc}\ds {\int }^{}{\text{cos}}^{3}\phantom{\rule{0.2em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds = - {\int }^{}{u}^{3}\phantom{\rule{0.2em}{0ex}}du\hfill \\[5mm]\ds \hfill &\ds =-\frac{1}{4}{u}^{4}+C\hfill \\[5mm]\ds \hfill &\ds =-\frac{1}{4}{\text{cos}}^{4}\phantom{\rule{0.2em}{0ex}}(x)+C.\hfill \end{array}

Evaluate \ds {\int }^{}{\text{sin}}^{4}(x)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Answer

\ds \frac{1}{5}{\text{sin}}^{5}(x)+C

Hint

Take \ds u=\text{sin}\phantom{\rule{0.1em}{0ex}}(x).

A Preliminary Example: Evaluating \ds  {\int }^{}{\text{cos}}^{j}(x)\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{k}(x)\phantom{\rule{0.2em}{0ex}}\,dx When \ds k is Odd

Evaluate \ds {\int }^{}{\text{cos}}^{2}(x)\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Solution

To convert this integral into a combination of integrals the form \ds {\int }^{}{\text{cos}}^{j}(x)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx , rewrite \ds {\text{sin}}^{3}(x)={\text{sin}}^{2}(x)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)=(1-{\text{cos}}^{2}(x))\sin(x).

We now make a substitution u=\cos(x), which means that du=-\sin(x)\,dx, and obtain

\ds \begin{array}{cc}\ds {\int }^{}{\text{cos}}^{2}(x)\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds ={\int }^{}{\text{cos}}^{2}(x)\left(1-{\text{cos}}^{2}(x)\right)\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill \\[5mm]\ds \hfill &\ds = - {\int }^{}{u}^{2}\left(1-{u}^{2}\right)du\hfill \\[5mm]\ds \hfill &\ds ={\int }^{}\left({u}^{4}-{u}^{2}\right)du\hfill \\[5mm]\ds \hfill &\ds =\frac{1}{5}{u}^{5}-\frac{1}{3}{u}^{3}+C\hfill \\[5mm]\ds \hfill &\ds =\frac{1}{5}{\text{cos}}^{5}(x)-\frac{1}{3}{\text{cos}}^{3}(x)+C.\hfill \end{array}

Evaluate \ds {\int }^{}{\text{cos}}^{3}(x)\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Answer

\ds \frac{1}{3}{\text{sin}}^{3}(x)-\frac{1}{5}{\text{sin}}^{5}(x)+C

Hint

Write \ds {\text{cos}}^{3}(x)={\text{cos}}^{2}(x)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)=\left(1-{\text{sin}}^{2}(x)\right)\text{cos}\phantom{\rule{0.1em}{0ex}}(x) and let \ds u=\text{sin}\phantom{\rule{0.1em}{0ex}}(x).

In the next example, we see the strategy that must be applied when there are only even powers of \ds \text{sin}\phantom{\rule{0.1em}{0ex}}(x) and \ds \text{cos}\phantom{\rule{0.1em}{0ex}}(x). For integrals of this type, the identities

\ds {\text{sin}}^{2}(x)=\frac{1}{2}-\frac{1}{2}\text{cos}\left(2x\right)=\frac{1-\text{cos}\left(2x\right)}{2}

and

\ds {\text{cos}}^{2}(x)=\frac{1}{2}+\frac{1}{2}\text{cos}\left(2x\right)=\frac{1+\text{cos}\left(2x\right)}{2}

are invaluable. These identities are sometimes known as power-reducing identities and they may be derived from the double-angle identity \ds \text{cos}\left(2x\right)={\text{cos}}^{2}(x)-{\text{sin}}^{2}(x) and the Pythagorean identity \ds {\text{cos}}^{2}(x)+{\text{sin}}^{2}(x)=1.

Integrating an Even Power of \ds  \text{sin}\phantom{\rule{0.1em}{0ex}}(x)

Evaluate \ds {\int }^{}{\text{sin}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Solution

To evaluate this integral, let’s use the trigonometric identity \ds {\text{sin}}^{2}(x)=\frac{1}{2}-\frac{1}{2}\text{cos}\left(2x\right). Thus,

\ds \begin{array}{cc}\ds {\int }^{}{\text{sin}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds ={\int }^{}\left(\frac{1}{2}-\frac{1}{2}\text{cos}\left(2x\right)\right)\,dx \hfill \\[5mm]\ds \hfill &\ds =\frac{1}{2}x-\frac{1}{4}\text{sin}\left(2x\right)+C.\hfill \end{array}

Evaluate \ds \int {\text{cos}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Answer

\ds \frac{1}{2}x+\frac{1}{4}\text{sin}\left(2x\right)+C

Hint

\ds {\text{cos}}^{2}(x)=\frac{1}{2}+\frac{1}{2}\text{cos}\left(2x\right)

The general process for integrating products of powers of \ds \text{sin}\phantom{\rule{0.1em}{0ex}}(x) and \ds \text{cos}\phantom{\rule{0.1em}{0ex}}(x) is summarized in the following set of guidelines.

Problem-Solving Strategy: Integrating Products of Powers of sin(x) and cos(x)

To evaluate \ds {\int }^{}{\text{cos}}^{j}(x)\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{k}(x)\phantom{\rule{0.2em}{0ex}}\,dx use the following strategies:

  1. If \ds k is odd, rewrite \ds {\text{sin}}^{k}(x)={\text{sin}}^{k-1}(x)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(x) and use the identity \ds {\text{sin}}^{2}(x)=1-{\text{cos}}^{2}(x) to rewrite \ds {\text{sin}}^{k-1}(x) in terms of \ds \text{cos}\phantom{\rule{0.1em}{0ex}}(x). Integrate using the substitution \ds u=\text{cos}\phantom{\rule{0.1em}{0ex}}(x). This substitution makes \ds du= - \text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx .
  2. If \ds j is odd, rewrite \ds {\text{cos}}^{j}(x)={\text{cos}}^{j-1}(x)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x) and use the identity \ds {\text{cos}}^{2}(x)=1-{\text{sin}}^{2}(x) to rewrite \ds {\text{cos}}^{j-1}(x) in terms of \ds \text{sin}\phantom{\rule{0.1em}{0ex}}(x). Integrate using the substitution \ds u=\text{sin}\phantom{\rule{0.1em}{0ex}}(x). This substitution makes \ds du=\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx .
    (Note: If both \ds j and \ds k are odd, either strategy 1 or strategy 2 may be used.)
  3. If both \ds j and \ds k are even, use identities \ds {\text{sin}}^{2}(x)=\frac12(1-\text{cos}\left(2x\right)) and \ds {\text{cos}}^{2}(x)=\frac12(1+\text{cos}\left(2x\right)). After applying these formulas, simplify and reapply strategies 2 and 3 to the combination of powers of \cos(2x) as appropriate.

Evaluating \ds  \int {\text{cos}}^{j}x\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{k}x\phantom{\rule{0.2em}{0ex}}\,dx When \ds j is Odd

Evaluate \ds {\int }^{}{\text{cos}}^{5}(x)\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{8}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Solution

Since the power on \ds \text{cos}\phantom{\rule{0.1em}{0ex}}(x) is odd, use strategy 2.

\ds \begin{array}{cccc}\hfill\ds{\int }^{}{\text{cos}}^{5}(x)\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{8}(x)\phantom{\rule{0.2em}{0ex}}\,dx &\ds ={\int }^{}{\text{cos}}^{4}(x)\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{8}(x)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds &\ds \text{Break off}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x).\hfill \\[5mm]\ds &\ds ={\int }^{}{\left({\text{cos}}^{2}(x)\right)}^{2}\sin^8(x)\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds &\ds \text{Rewrite}\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{4}(x)={\left({\text{cos}}^{2}(x)\right)}^{2}.\hfill \\[5mm]\ds &\ds ={\int }^{}{\left(1-{\text{sin}}^{2}(x)\right)}^{2}\sin^8(x)\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds &\ds \text{Substitute}\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{2}(x)=1-{\text{sin}}^{2}(x).\hfill \\[5mm]\ds &\ds ={\int }^{}{\left(1-{u}^{2}\right)}^{2}u^8\,du\hfill &\ds &\ds \text{Let}\phantom{\rule{0.2em}{0ex}}u=\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}du= \text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx .\hfill \\[5mm]\ds &\ds ={\int }^{}\left( {u}^{8}-2{u}^{10}+{u}^{12}\right)du\hfill &\ds &\ds \text{Expand}.\hfill \\[5mm]\ds &\ds =\frac{1}{9}{u}^{9}-\frac{2}{11}{u}^{11}+\frac{1}{13}{u}^{13}+C\hfill &\ds &\ds \text{Evaluate the integral}.\hfill \\[5mm]\ds &\ds =\frac{1}{9}{\text{sin}}^{9}(x)-\frac{2}{11}{\text{sin}}^{11}(x)+\frac{1}{13}{\text{sin}}^{13}(x)+C.\hfill &\ds &\ds \text{Substitute}\phantom{\rule{0.2em}{0ex}}u=\text{sin}\phantom{\rule{0.1em}{0ex}}(x).\hfill \end{array}

Evaluating \ds{\int }^{}{\text{cos}}^{j}(x)\phantom{\rule{0.1em}{0ex}}{\text{sin}}^{k}(x)\phantom{\rule{0.2em}{0ex}}\,dx When \ds k and \ds j are Even

Evaluate \ds {\int }^{}{\text{sin}}^{4}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Solution

Since both the powers of \ds \text{sin}\phantom{\rule{0.1em}{0ex}}(x)  and \ds \text{cos}\phantom{\rule{0.1em}{0ex}}(x) are even \ds \left(k=4,\ j=0\right), we must use strategy 3. Thus,

\ds \begin{array}{cccc}\hfill \ds{\int }^{}{\text{sin}}^{4}(x)\phantom{\rule{0.2em}{0ex}}\,dx &\ds ={\int }^{}{\left({\text{sin}}^{2}(x)\right)}^{2}\,dx \hfill &\ds &\ds \text{Rewrite}\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{4}(x)={\left({\text{sin}}^{2}(x)\right)}^{2}.\hfill \\[5mm]\ds &\ds ={\int }^{}{\left(\frac{1}{2}-\frac{1}{2}\text{cos}\left(2x\right)\right)}^{2}\,dx \hfill &\ds &\ds \text{Substitute}\phantom{\rule{0.2em}{0ex}}{\text{sin}}^{2}(x)=\frac{1}{2}-\frac{1}{2}\text{cos}\left(2x\right).\hfill \\[5mm]\ds &\ds ={\int }^{}\left(\frac{1}{4}-\frac{1}{2}\text{cos}\left(2x\right)+\frac{1}{4}{\text{cos}}^{2}\left(2x\right)\right)\,dx \hfill &\ds &\ds \text{Expand}\phantom{\rule{0.1em}{0ex}}{\left(\frac{1}{2}-\frac{1}{2}\text{cos}\left(2x\right)\right)}^{2}.\hfill \ds &\ds \end{array}

Since \ds {\text{cos}}^{2}\left(2x\right) has an even power, we use strategy 3 again and substitute \ds {\text{cos}}^{2}\left(2x\right)=\frac{1}{2}+\frac{1}{2}\text{cos}\left(4x\right) to continue the equalities as follows:

\ds \begin{array}{cc}\ds={\int }^{}\left(\frac{1}{4}-\frac{1}{2}\text{cos}\left(2x\right)+\frac{1}{4}\left(\frac{1}{2}+\frac{1}{2}\text{cos}\left(4x\right)\right)\right)\,dx \hfill &\hfill\\[5mm]\ds={\int }^{}\left(\frac{3}{8}-\frac{1}{2}\text{cos}\left(2x\right)+\frac{1}{8}\text{cos}\left(4x\right)\right)\,dx \hfill &\ds \text{Simplify}.\hfill \\[5mm]\ds =\frac{3}{8}x-\frac{1}{4}\text{sin}\left(2x\right)+\frac{1}{32}\phantom{\rule{0.1em}{0ex}}\text{sin}\left(4x\right)+C.\hfill &\ds \text{Evaluate the integral}.\hfill \end{array}

Evaluate \ds {\int }^{}{\text{cos}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Answer

\ds \text{sin}\phantom{\rule{0.1em}{0ex}}(x)-\frac{1}{3}{\text{sin}}^{3}(x)+C

Hint

Use strategy 2. Write \ds {\text{cos}}^{3}(x)={\text{cos}}^{2}(x)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x) and substitute \ds {\text{cos}}^{2}(x)=1-{\text{sin}}^{2}(x).

Evaluate \ds {\int }^{}{\text{cos}}^{2}\left(3x\right)\,dx .

Answer

\ds \frac{1}{2}x+\frac{1}{12}\phantom{\rule{0.1em}{0ex}}\text{sin}\left(6x\right)+C

Hint

Use strategy 3 and substitute \ds {\text{cos}}^{2}\left(3x\right)=\frac{1}{2}+\frac{1}{2}\text{cos}\left(6x\right).

In some areas of physics, such as quantum mechanics, signal processing, and the computation of Fourier series, it is often necessary to integrate products that include \ds \text{sin}\left(ax\right), \ds \text{sin}\left(bx\right), \ds \text{cos}\left(ax\right), and \ds \text{cos}\left(bx\right). These integrals are evaluated by applying trigonometric identities, as outlined below.

Integrating Products of Sines and Cosines of Different Angles

To integrate products involving \ds \text{sin}\left(ax\right), \ds \text{sin}\left(bx\right), \ds \text{cos}\left(ax\right), and \ds \text{cos}\left(bx\right), use the following identities

\ds \begin{array}{ll}\text{sin}\left(ax\right)\text{sin}\left(bx\right)&=\ds\frac{1}{2}\text{cos}\big(\left(a-b\right)x\big)-\frac{1}{2}\text{cos}\big(\left(a+b\right)x\big)\hfill\\[5mm]\ds \text{sin}\left(ax\right)\text{cos}\left(bx\right) &\ds=\frac{1}{2}\text{sin}\big(\left(a-b\right)x\big)+\frac{1}{2}\text{sin}\big(\left(a+b\right)x\big)\hfill\\[5mm]\ds \text{cos}\left(ax\right)\text{cos}\left(bx\right)&\ds=\frac{1}{2}\text{cos}\big(\left(a-b\right)x\big)+\frac{1}{2}\text{cos}\big(\left(a+b\right)x\big)\hfill\end{array}

These formulas may be derived from the sum-of-angle formulas for sine and cosine.

Evaluating \ds  \int \text{sin}\left(ax\right)\text{cos}\left(bx\right)\,dx

Evaluate \ds {\int }^{}\text{sin}\left(5x\right)\text{cos}\left(3x\right)\,dx .

Solution

Apply the identity \ds \text{sin}\left(5x\right)\text{cos}\left(3x\right)=\frac{1}{2}\text{sin}\left(2x\right)+\frac{1}{2}\text{sin}\left(8x\right). Thus,

\ds \begin{array}{cc}\ds {\int }^{}\text{sin}\left(5x\right)\text{cos}\left(3x\right)\,dx \hfill &\ds ={\int }^{}\left(\frac{1}{2}\text{sin}\left(2x\right)+\frac{1}{2}\text{sin}\left(8x\right)\right)\,dx \hfill \\[5mm]\ds \hfill &\ds =-\frac{1}{4}\text{cos}\left(2x\right)-\frac{1}{16}\phantom{\rule{0.1em}{0ex}}\text{cos}\left(8x\right)+C.\hfill \end{array}

Evaluate \ds {\int }^{}\text{cos}\left(6x\right)\text{cos}\left(5x\right)\,dx .

Answer

\ds \frac{1}{2}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)+\frac{1}{22}\phantom{\rule{0.1em}{0ex}}\text{sin}\left(11x\right)+C

Hint

Rewrite \ds \text{cos}\left(6x\right)\text{cos}\left(5x\right)=\frac{1}{2}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)+\frac{1}{2}\text{cos}\left(11x\right).

Integrating Products and Powers of tan(x) and sec(x)

Before discussing the integration of products of powers of \ds \text{tan}\phantom{\rule{0.1em}{0ex}}(x) and \ds \text{sec}\phantom{\rule{0.1em}{0ex}}(x), it is useful to recall the integrals involving \ds \text{tan}\phantom{\rule{0.1em}{0ex}}(x) and \ds \text{sec}\phantom{\rule{0.1em}{0ex}}(x) we have already learned:

  1. \ds {\int }^{}{\text{sec}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx =\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+C,
  2. \ds {\int }^{}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx =\text{sec}\phantom{\rule{0.1em}{0ex}}(x)+C,
  3. \ds {\int }^{}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx =\text{ln}|\text{sec}\phantom{\rule{0.1em}{0ex}}(x)|+C,
  4. \ds {\int }^{}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx =\text{ln}|\text{sec}\phantom{\rule{0.1em}{0ex}}(x)+\text{tan}\phantom{\rule{0.1em}{0ex}}(x)|+C.

(Formulas 1 and 2 come directly from the table of indefinite integrals, while formulas 3 and 4 were derived in Section 1.5 Substitution.)

For most integrals of products of powers of \ds \text{tan}\phantom{\rule{0.1em}{0ex}}(x) and \ds \text{sec}\phantom{\rule{0.1em}{0ex}}(x), we rewrite the expression we wish to integrate as the sum or difference of integrals of the form \ds {\int }^{}{\text{tan}}^{j}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx or \ds {\int }^{}{\text{sec}}^{j}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx . As we see in the following example, we can evaluate these new integrals by using appropriate substitution.

Evaluating \ds  {\int }^{}{\text{sec}}^{j}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Evaluate \ds {\int }^{}{\text{sec}}^{5}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Solution

Start by rewriting \ds {\text{sec}}^{5}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x) as \ds {\text{sec}}^{4}(x)\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x). If we now let u=\sec(x), then du=\sec(x)\tan(x)\,dx, and so

\ds \begin{array}{cc}\ds{\int }^{}{\text{sec}}^{5}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}dx &\ds ={\int }^{}{\text{sec}}^{4}(x)\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}dx \hfill \\[5mm]\ds &\ds ={\int }^{}{u}^{4}\,du\hfill \\[5mm]\ds &\ds =\frac{1}{5}{u}^{5}+C\hfill \\[5mm]\ds &\ds =\frac{1}{5}{\text{sec}}^{5}(x)+C.\hfill \end{array}

Evaluate \ds {\int }^{}{\text{tan}}^{5}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Answer

\ds \frac{1}{6}{\text{tan}}^{6}(x)+C

Hint

Let \ds u=\text{tan}\phantom{\rule{0.1em}{0ex}}(x) and \ds du={\text{sec}}^{2}(x).

We now take a look at the various strategies for integrating products of powers of \ds \text{sec}\phantom{\rule{0.1em}{0ex}}(x) and \ds \text{tan}\phantom{\rule{0.1em}{0ex}}(x).

Problem-Solving Strategy: Evaluating \ds {\int }^{}{\text{tan}}^{k}x\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{j}x\phantom{\rule{0.2em}{0ex}}\,dx

To evaluate \ds {\int }^{}{\text{tan}}^{k}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{j}(x)\phantom{\rule{0.2em}{0ex}}\,dx , use the following strategies:

  1. If \ds j is even and \ds j\ge 2, rewrite \ds {\text{sec}}^{j}(x)={\text{sec}}^{j-2}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}(x) and use \ds {\text{sec}}^{2}(x)={\text{tan}}^{2}(x)+1 to rewrite \ds {\text{sec}}^{j-2}(x) in terms of \ds \text{tan}\phantom{\rule{0.1em}{0ex}}(x). Let \ds u=\text{tan}\phantom{\rule{0.1em}{0ex}}(x) and \ds du={\text{sec}}^{2}(x).
  2. If \ds k is odd and \ds j\ge 1, rewrite \ds {\text{tan}}^{k}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{j}(x)={\text{tan}}^{k-1}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{j-1}(x)\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x) and use \ds {\text{tan}}^{2}(x)={\text{sec}}^{2}(x)-1 to express \ds {\text{tan}}^{k-1}(x) in terms of \ds \text{sec}\phantom{\rule{0.1em}{0ex}}(x). Let \ds u=\text{sec}\phantom{\rule{0.1em}{0ex}}(x) and \ds du=\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx .
    (Note: If \ds j is even and \ds k is odd, then either strategy 1 or strategy 2 may be used.)
  3. If \ds k\ge3 is odd and \ds j=0, rewrite \begin{array}{cc}\ds {\text{tan}}^{k}(x)&={\text{tan}}^{k-2}(x)\phantom{\rule{0.1em}{0ex}}{\text{tan}}^{2}(x)={\text{tan}}^{k-2}(x)\left({\text{sec}}^{2}(x)-1\right)\hfill\\[3mm]\ds&={\text{tan}}^{k-2}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}(x)-{\text{tan}}^{k-2}(x).\hfill\\&\hfill\end{array}
    It may be necessary to repeat this process on the \ds {\text{tan}}^{k-2}(x) term.
  4. If \ds k is even and \ds j is odd, then use \ds {\text{tan}}^{2}(x)={\text{sec}}^{2}(x)-1 to express \ds {\text{tan}}^{k}(x) in terms of \ds \text{sec}\phantom{\rule{0.1em}{0ex}}(x). Use integration by parts to integrate odd powers of \ds \text{sec}\phantom{\rule{0.1em}{0ex}}(x).

Evaluating \ds {\int }^{}{\text{tan}}^{k}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{j}(x)\phantom{\rule{0.2em}{0ex}}\,dx When j is Even

Evaluate \ds {\int }^{}{\text{tan}}^{6}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{4}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Solution

Since the power on \ds \text{sec}\phantom{\rule{0.1em}{0ex}}(x) is even, rewrite \ds {\text{sec}}^{4}(x)={\text{sec}}^{2}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}(x) and use \ds {\text{sec}}^{2}(x)={\text{tan}}^{2}(x)+1 to express the first \ds {\text{sec}}^{2}(x) in terms of \ds \text{tan}\phantom{\rule{0.1em}{0ex}}(x). We now make a substitution u=\tan(x), in which case du=\sec^2(x)\,dx, and we obtain

\ds \begin{array}{cc}\ds{\int }^{}{\text{tan}}^{6}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{4}(x)\phantom{\rule{0.2em}{0ex}}\,dx &\ds ={\int }^{}{\text{tan}}^{6}(x)\left({\text{tan}}^{2}(x)+1\right){\text{sec}}^{2}(x)\hfill \\[5mm]\ds &\ds ={\int }^{}{u}^{6}\left({u}^{2}+1\right)du\hfill \\[5mm]\ds &\ds ={\int }^{}\left({u}^{8}+{u}^{6}\right)du\hfill \\[5mm]\ds &\ds =\frac{1}{9}{u}^{9}+\frac{1}{7}{u}^{7}+C\hfill \\[5mm]\ds &\ds =\frac{1}{9}{\text{tan}}^{9}(x)+\frac{1}{7}{\text{tan}}^{7}(x)+C.\hfill &\ds &\ds \hfill \end{array}

Evaluating \ds {\int }^{}{\text{tan}}^{k}x\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{j}x\phantom{\rule{0.2em}{0ex}}\,dx When k is Odd

Evaluate \ds {\int }^{}{\text{tan}}^{5}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Solution

Since the power of \ds \text{tan}\phantom{\rule{0.1em}{0ex}}(x) is odd, we begin by rewriting \ds {\text{tan}}^{5}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{3}(x)={\text{tan}}^{4}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}(x)\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x). We then notice that \tan^4(x)=(\tan^2(x))^2=(\sec^2(x)-1)^2, and make a substitution u=\sec(x) with du=\sec(x)\tan(x)\,dx. With this, we obtain

\ds \begin{array}{cc}\ds \hfill {\int }^{}{\text{tan}}^{5}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx &\ds =\ds {\int }^{}{\left(\sec^2(x)-1\right)}^{2}{\text{sec}}^{2}(x)\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill \\[5mm]\ds &\ds = {\int }^{}{\left({u}^{2}-1\right)}^{2}{u}^{2}du\hfill \\[5mm]\hfill &=\ds {\int }^{}\left({u}^{6}-2{u}^{4}+{u}^{2}\right)du\hfill \\[5mm]\ds &\ds = \frac{1}{7}{u}^{7}-\frac{2}{5}{u}^{5}+\frac{1}{3}{u}^{3}+C\hfill  \\[5mm]\ds &\ds =\frac{1}{7}{\text{sec}}^{7}(x)-\frac{2}{5}{\text{sec}}^{5}(x)+\frac{1}{3}{\text{sec}}^{3}(x)+C.\hfill \end{array}

Evaluating \ds {\int }^{}{\text{tan}}^{k}x\phantom{\rule{0.2em}{0ex}}\,dx When k\ge3 is Odd

Evaluate \ds {\int }^{}{\text{tan}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Solution

We begin by rewriting \begin{array}{cc}\ds {\text{tan}}^{3}(x)&={\text{tan}}(x)\phantom{\rule{0.1em}{0ex}}{\text{tan}}^{2}(x)={\text{tan}}(x)\left({\text{sec}}^{2}(x)-1\right)\hfill\\[3mm]\ds&={\text{tan}}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}(x)-{\text{tan}}(x).\hfill\\&\hfill\end{array}
Therefore,

\ds \begin{array}{cc}\ds {\int }^{}{\text{tan}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds ={\int }^{}\left(\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}(x)-\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\right)\,dx \hfill \\[5mm]\ds \hfill &\ds ={\int }^{}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx -{\int }^{}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill \\[5mm]\ds \hfill &\ds =\frac{1}{2}{\text{tan}}^{2}(x)-\text{ln}|\text{sec}\phantom{\rule{0.1em}{0ex}}(x)|+C.\hfill \end{array}

For the first integral, we used the substitution \ds u=\text{tan}\phantom{\rule{0.1em}{0ex}}(x), and for the second integral, we used the formula.

Evaluating \ds {\int }^{}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Integrate \ds {\int }^{}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Solution

This integral requires integration by parts. Let \ds u=\text{sec}\phantom{\rule{0.1em}{0ex}}(x) and \ds dv={\text{sec}}^{2}(x)\,dx. These choices make \ds du=\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\,dx and \ds v=\text{tan}\phantom{\rule{0.1em}{0ex}}(x). Thus,

\ds \begin{array}{cccc}\hfill \ds{\int }^{}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx &\ds =\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)-{\int }^{}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds &\ds \\[5mm]\ds &\ds =\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)-{\int }^{}{\text{tan}}^{2}(x)\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds &\ds \text{Simplify}.\hfill \\[5mm]\ds &\ds =\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)-{\int }^{}\left({\text{sec}}^{2}(x)-1\right)\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds &\ds \text{Substitute}\phantom{\rule{0.2em}{0ex}}{\text{tan}}^{2}(x)={\text{sec}}^{2}(x)-1.\hfill \\[5mm]\ds &\ds =\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+{\int }^{}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx -{\int }^{}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds &\ds \text{Rewrite}.\hfill \\[5mm]\ds &\ds =\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+\text{ln}|\text{sec}\phantom{\rule{0.1em}{0ex}}(x)+\text{tan}\phantom{\rule{0.1em}{0ex}}(x)|-{\int }^{}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx .\hfill &\ds &\ds \text{Evaluate}{\int }^{}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx .\hfill \end{array}

We now have

\ds {\int }^{}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx =\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+\text{ln}|\text{sec}\phantom{\rule{0.1em}{0ex}}(x)+\text{tan}\phantom{\rule{0.1em}{0ex}}(x)|-{\int }^{}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx .
We see that the last integral is the same as the original one. Let I be a particular antiderivative of \ds\sec^3(x). Substituting I instead of \ds {\int }^{\text{​}}\sec^3(x)\,dx into the above equality, we get the following equation for I.
\ds I=\sec(x)\tan(x)+\ln|\sec(x)+\tan(x)|-I.

Adding I to both sides, we obtain

\ds 2I =\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+\text{ln}|\text{sec}\phantom{\rule{0.1em}{0ex}}(x)+\text{tan}\phantom{\rule{0.1em}{0ex}}(x)|.

Dividing by 2, we arrive at

\ds I =\frac{1}{2}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+\frac{1}{2}\text{ln}|\text{sec}\phantom{\rule{0.1em}{0ex}}(x)+\text{tan}\phantom{\rule{0.1em}{0ex}}(x)|.
Since an indefinite integral can be found by adding an arbitrary constant to a particular antiderivative of the integrand, we obtain that \ds\int\sec^3(x)\,dx=I+C=\frac{1}{2}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+\frac{1}{2}\text{ln}|\text{sec}\phantom{\rule{0.1em}{0ex}}(x)+\text{tan}\phantom{\rule{0.1em}{0ex}}(x)|+C.

Evaluate \ds {\int }^{}{\text{tan}}^{3}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{7}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Answer

\ds \frac{1}{9}{\text{sec}}^{9}(x)-\frac{1}{7}{\text{sec}}^{7}(x)+C

Reduction Formulas

Evaluating \ds {\int }^{}{\text{sec}}^{n}(x)\phantom{\rule{0.2em}{0ex}}\,dx for odd values of \ds n requires integration by parts. In addition, we must also know the value of \ds {\int }^{}{\text{sec}}^{n-2}(x)\phantom{\rule{0.2em}{0ex}}\,dx to evaluate \ds {\int }^{}{\text{sec}}^{n}(x)\phantom{\rule{0.2em}{0ex}}\,dx . The evaluation of \ds {\int }^{}{\text{tan}}^{n}(x)\phantom{\rule{0.2em}{0ex}}\,dx also requires being able to integrate \ds {\int }^{}{\text{tan}}^{n-2}(x)\phantom{\rule{0.2em}{0ex}}\,dx . To make the process easier, we can derive and apply the following power reduction formulas. They allow us to replace the integral of a power of \ds \text{sec}\phantom{\rule{0.1em}{0ex}}(x) or \ds \text{tan}\phantom{\rule{0.1em}{0ex}}(x) with the integral of a lower power of \ds \text{sec}\phantom{\rule{0.1em}{0ex}}(x) or \ds \text{tan}\phantom{\rule{0.1em}{0ex}}(x).

Reduction Formulas for \ds {\int }^{}{\text{sec}}^{n}(x)\phantom{\rule{0.2em}{0ex}}\,dx and \ds {\int }^{}{\text{tan}}^{n}(x)\phantom{\rule{0.2em}{0ex}}\,dx

\ds \phantom{\rule{4.08em}{0ex}}{\int }^{}{\text{sec}}^{n}(x)\phantom{\rule{0.2em}{0ex}}\,dx =\frac{1}{n-1}{\text{sec}}^{n-2}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+\frac{n-2}{n-1}{\int }^{}{\text{sec}}^{n-2}(x)\phantom{\rule{0.2em}{0ex}}\,dx
\ds {\int }^{}{\text{tan}}^{n}(x)\phantom{\rule{0.2em}{0ex}}\,dx =\frac{1}{n-1}{\text{tan}}^{n-1}(x)-{\int }^{}{\text{tan}}^{n-2}(x)\phantom{\rule{0.2em}{0ex}}\,dx

The first power reduction rule may be verified by applying integration by parts. The second may be verified by following the strategy outlined for integrating odd powers of \ds \text{tan}\phantom{\rule{0.1em}{0ex}}(x).

Revisiting \ds {\int }^{}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Apply a reduction formula to evaluate \ds {\int }^{}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Solution

By applying the first reduction formula with n=3, we obtain

\ds \begin{array}{cc}\ds {\int }^{}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds =\frac{1}{3-1}\text{sec}^{3-2}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+\frac{3-2}{3-1}{\int }^{}\text{sec}^{3-2}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill \\[5mm]\ds \hfill &\ds =\frac{1}{2}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+\frac{1}{2}\text{ln}|\text{sec}\phantom{\rule{0.1em}{0ex}}(x)+\text{tan}\phantom{\rule{0.1em}{0ex}}(x)|+C.\hfill \end{array}

Using a Reduction Formula

Evaluate \ds {\int }^{}{\text{tan}}^{4}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Solution

Applying the second reduction formula with n=4, we obtain

\ds{\int }^{}{\text{tan}}^{4}(x)\phantom{\rule{0.2em}{0ex}}\,dx \ds =\frac{1}{4-1}{\text{tan}}^{4-1}(x)-{\int }^{}{\text{tan}}^{4-2}(x)\phantom{\rule{0.2em}{0ex}}\,dx.

To evaluate \int \tan^2(x)\,dx, we apply the second reduction formula with n=2, which allows to continue the chain of equalities as follows:

\ds \begin{array}{cc}\hfill \ds{\int }^{}{\text{tan}}^{4}(x)\phantom{\rule{0.2em}{0ex}}\,dx &\ds =\frac{1}{3}{\text{tan}}^{3}(x)-{\int }^{}{\text{tan}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill \\[5mm]\ds &\ds =\frac{1}{3}{\text{tan}}^{3}(x)-\left(\frac1{2-1}\text{tan}^{2-1}\phantom{\rule{0.1em}{0ex}}(x)-{\int }^{}{\text{tan}}^{2-2}(x)\phantom{\rule{0.2em}{0ex}}\,dx \right)\hfill\\[5mm]\ds &\ds =\frac{1}{3}{\text{tan}}^{3}(x)-\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+{\int }^{}1\phantom{\rule{0.2em}{0ex}}\,dx \hfill  \\[5mm]\ds &\ds =\frac{1}{3}{\text{tan}}^{3}(x)-\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+x+C.\hfill\end{array}

Apply the reduction formula to \ds {\int }^{}{\text{sec}}^{5}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Answer

\ds {\int }^{}{\text{sec}}^{5}(x)\phantom{\rule{0.2em}{0ex}}\,dx =\frac{1}{4}{\text{sec}}^{3}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)-\frac{3}{4}{\int }^{}{\text{sec}}^{3}(x)\,dx

Key Concepts

Integrals of trigonometric functions can be evaluated using various strategies. These strategies include the following.

  1. Applying trigonometric identities to rewrite the integrand so that it may be evaluated via an apropriate substitution.
  2. Using integration by parts.
  3. Applying trigonometric identities to rewrite products of sines and cosines with different arguments as the sum of individual sine and cosine functions.
  4. Applying reduction formulas.

Key Equations

  • Sine Products
    \ds \text{sin}\left(ax\right)\text{sin}\left(bx\right)=\frac{1}{2}\text{cos}\left(\left(a-b\right)x\right)-\frac{1}{2}\text{cos}\left(\left(a+b\right)x\right)
  • Sine and Cosine Products
    \ds \text{sin}\left(ax\right)\text{cos}\left(bx\right)=\frac{1}{2}\text{sin}\left(\left(a-b\right)x\right)+\frac{1}{2}\text{sin}\left(\left(a+b\right)x\right)
  • Cosine Products
    \ds \text{cos}\left(ax\right)\text{cos}\left(bx\right)=\frac{1}{2}\text{cos}\left(\left(a-b\right)x\right)+\frac{1}{2}\text{cos}\left(\left(a+b\right)x\right)
  • Power Reduction Formula
    \ds {\int }^{}{\text{sec}}^{n}(x)\phantom{\rule{0.2em}{0ex}}\,dx =\frac{1}{n-1}{\text{sec}}^{n-1}(x)+\frac{n-2}{n-1}{\int }^{}{\text{sec}}^{n-2}(x)\phantom{\rule{0.2em}{0ex}}\,dx
  • Power Reduction Formula
    \ds {\int }^{}{\text{tan}}^{n}(x)\phantom{\rule{0.2em}{0ex}}\,dx =\frac{1}{n-1}{\text{tan}}^{n-1}(x)-{\int }^{}{\text{tan}}^{n-2}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Exercises

Fill in the blank to make a true statement.

1. \ds {\text{sin}}^{2}(x)+\underline{\phantom{aaaaaaaaa}}=1

Answer

\ds {\text{cos}}^{2}(x)

2. \ds {\text{sec}}^{2}(x)-1=\underline{\phantom{aaaaaaaaa}}

Use an identity to reduce the power of the trigonometric function to a trigonometric function raised to the first power.

3. \ds {\text{sin}}^{2}(x)=\underline{\phantom{aaaaaaaaa}}

Answer

\ds \frac{1-\text{cos}\left(2x\right)}{2}

4. \ds {\text{cos}}^{2}(x)=\underline{\phantom{aaaaaaaaa}}

Evaluate each of the following integrals using appropriate substitution.

5. \ds \int {\text{sin}}^{3}(x)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Answer

\ds \frac{{\text{sin}}^{4}(x)}{4}+C

6. \ds \int \sqrt{\text{cos}\phantom{\rule{0.1em}{0ex}}(x)}\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

7. \ds \int {\text{tan}}^{5}\left(2x\right){\text{sec}}^{2}\left(2x\right)\,dx

Answer

\ds \frac{1}{12}{\text{tan}}^{6}\left(2x\right)+C

8. \ds \int \frac{\text{cos}\left(2x\right)}{{\text{sin}}^{7}\left(2x\right)}\,dx

9. \ds \int \text{tan}\left(\frac{x}{2}\right){\text{sec}}^{4}\left(\frac{x}{2}\right)\,dx

Answer

\ds \frac12{\text{sec}}^{4}\left(\frac{x}{2}\right)+C

10. \ds \int \frac{{\text{sec}}^{2}(x)\phantom{\rule{0.2em}{0ex}}}{\sqrt{{\text{tan}}(x)}\phantom{\rule{0.1em}{0ex}}}\,dx

Compute the following integrals using the guidelines for integrating powers of trigonometric functions. (Note: Some of the problems may be done using techniques of integration learned previously.)

11. \ds \int {\text{sin}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Answer

\ds - \text{cos}\phantom{\rule{0.1em}{0ex}}(x)+\frac{{\text{cos}}^{3}(x)}{3}+C=-\frac{3\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)}{4}+\frac{1}{12}\phantom{\rule{0.1em}{0ex}}\text{cos}\left(3x\right)+C

12. \ds \int {\text{cos}}^{5}(x)\phantom{\rule{0.2em}{0ex}}\,dx

13. \ds \int \frac{\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}}{\text{cos}^3\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}}\,dx

Answer

\ds \frac{1}{2\text{cos}^{2}(x)}+C

14. \ds \int \frac{\cos^3(x)}{\sin(x)}\,dx

15. \ds \int {\text{sin}}^{5}(x)\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Answer

\ds -\frac17\cos^7(x)+\frac25\cos^5(x)-\frac13\cos^3(x)+C

16. \ds \int {\text{sin}}^{3}(x)\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx

17. \ds \int \sqrt{\text{sin}\phantom{\rule{0.1em}{0ex}}(x)}\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Answer

\ds \frac{2}{3}{\left(\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\right)}^{3\text{/}2}+C

18. \ds \int \sqrt{\text{sin}\phantom{\rule{0.1em}{0ex}}(x)}\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx

19. \ds \int {\text{tan}}^{2}(x)\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

(Hint: use integration by parts)

Answer

\ds \frac{1}{2}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)-\frac{1}{2}\text{ln}\left(\text{sec}\phantom{\rule{0.1em}{0ex}}(x)+\text{tan}\phantom{\rule{0.1em}{0ex}}(x)\right)+C

20. \ds \int \text{tan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{3}(x)\phantom{\rule{0.2em}{0ex}}\,dx

21. \ds \int {\text{sec}}^{4}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Answer

\ds \text{tan}\phantom{\rule{0.1em}{0ex}}(x)+\frac{{\text{tan}}^{3}(x)}{3}+C=\frac{2\phantom{\rule{0.1em}{0ex}}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)}{3}+\frac{1}{3}\text{sec}^{2}{\left(x\right)}\text{tan}\phantom{\rule{0.1em}{0ex}}(x)+C

22. \ds \int \text{cot}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\csc^5(x)\,dx

23. \ds \int \sqrt[3]{\cot^5(x)}\text{csc}^4\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Answer

\ds - \frac3{14}(\cot(x))^{14\text{/}3}-\frac38(\cot(x))^{8\text{/}3}+C

24. \ds \int \frac{{\text{tan}}^{3}(x)}{\text{sec}\phantom{\rule{0.1em}{0ex}}(x)}\,dx

For the following exercises, evaluate the integrals involving parameter a.

25. \ds \int {\text{sin}}^{2}(ax)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(ax)\phantom{\rule{0.2em}{0ex}}\,dx

Answer

\ds \frac{{\text{sin}}^{3}\left(ax\right)}{3a}+C

26. \ds \int \text{sin}\phantom{\rule{0.1em}{0ex}}(ax)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(2ax)\phantom{\rule{0.2em}{0ex}}\,dx .

Use the double-angle formulas to evaluate the following integrals.

27. \ds \int\limits_{0}^{\pi }{\text{sin}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Answer

\ds \frac{\pi }{2}

28. \ds \int\limits_{0}^{\pi\text{/}2 }{\text{cos}}^{4}(x)\phantom{\rule{0.2em}{0ex}}\,dx

29. \ds \int {\text{cos}}^{2}(5x+1)\phantom{\rule{0.2em}{0ex}}\,dx

Answer

\ds \frac{x}{2}+\frac{1}{20}\phantom{\rule{0.1em}{0ex}}\text{sin}\left(10x+2\right)+C

30. \ds \int {\text{sin}}^{2}(x)\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx

31. \ds \int {\text{sin}}(2x)\phantom{\rule{0.2em}{0ex}}\cos^2(x)\,dx

Answer

\ds -\frac14\cos(2x)+\frac18\sin^2(2x)+C=-\frac12\cos^4(x)+C

32. \ds \int \frac{{\text{cos}}^{2}(x)\phantom{\rule{0.1em}{0ex}}}{{\text{cos}}\left(2x\right)}\,dx

(Hint: to integrate \ds\frac1{\cos(2x)}, rewrite it as \sec(2x) and use the formula.)

For the following exercises, evaluate the definite integrals using an appropriate trigonometric formula.

33. \ds \int\limits_{\pi\text{/}3}^{\pi\text{/}2 }\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(2x)\phantom{\rule{0.2em}{0ex}}\,dx

Answer

\ds\frac1{12}

34. \ds \int\limits_{0}^{\pi }\text{sin}\phantom{\rule{0.1em}{0ex}}(3x)\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(5x)\phantom{\rule{0.2em}{0ex}}\,dx

35. \ds \int\limits_{-\pi\text{/}2}^{0}\text{cos}\left(99x\right)\text{sin}\left(101x\right)\,dx

Answer

\ds-\frac12

36. \ds \int\limits_{0}^{\pi\text{/}6 }{\text{cos}}\left(3x\right)\cos(6x)\,dx

37. \ds \int\limits_{0}^{\pi/4 }\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.1em}{0ex}}\text{sin}\left(2x\right)\text{sin}\left(3x\right)\,dx

Answer

\ds\frac5{24}

38. \ds \int\limits_{0}^{\pi }\text{cos}\left(\frac x2\right)\text{sin}\left(\frac {3x}2\right)\,dx

39. \ds \int\limits_{\pi \text{/}6}^{\pi \text{/}2}\sin^2(x)\sin(3x)\,dx

Answer

\ds-\frac{\sqrt3}{10}

40. \ds \int\limits_{ - \pi \text{/}3}^{0}\sqrt{{\text{sec}}^{2}(x)-1}\phantom{\rule{0.1em}{0ex}}\,dx

41. \ds \int\limits_{0}^{\pi \text{/}2}\sqrt{1-\text{cos}\left(2x\right)}\phantom{\rule{0.1em}{0ex}}\,dx

Answer

\ds \sqrt{2}

42. Find the area of the region bounded by the curves \ds y=\text{sin}\phantom{\rule{0.1em}{0ex}}(x),y={\text{sin}}^{3}(x),x=0,\text{and}\phantom{\rule{0.2em}{0ex}}x=\frac{\pi }{2}.

43. Find the area of the region bounded by the curves \ds y={\text{cos}}^{2}(x),y={\text{sin}}^{2}(x),x=-\frac{\pi }{4},\text{and}\phantom{\rule{0.2em}{0ex}}x=\frac{\pi }{4}.

Answer

1

44. A particle moves in a straight line with the velocity function \ds v\left(t\right)=\text{sin}\left(\omega t\right){\text{cos}}^{2}\left(\omega t\right). Find its position function \ds x=f\left(t\right) if \ds f\left(0\right)=0.

45. Find the average value of the function \ds f\left(x\right)={\text{sin}}^{2}(x)\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{3}(x) over the interval \ds \left[ -\frac{\pi}2 ,0\right].

Answer

\ds\frac4{15\pi}

46. Find the length of the curve \ds y=\text{ln}\left(\text{csc}\phantom{\rule{0.1em}{0ex}}(x)\right),\frac{\pi }{4}\le x\le \frac{\pi }{2}.

47. Find the length of the curve \ds y=\text{ln}\left(\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\right),\frac{\pi }{3}\le x\le \frac{\pi }{2}.

Answer

\ds \frac12\text{ln}\left(3\right)

48. Let R be the region below the curve \ds y=\text{cos}\left(3x\right) and above the x-axis over the interval \ds\left[0,\frac{\pi}{36}\right]. Find the volume generated by revolving R about the x-axis.

49. The inner product of two functions f and g over \ds \left[a,b\right] is defined by \ds f\left(x\right)\cdot g\left(x\right)=\langle f,g\rangle=\int\limits_{a}^{b}f(x)g(x)\,dx . Two non-zero functions f and g are said to be orthogonal if \ds \langle f,g\rangle=0.

Show that \ds \text{sin}\left(2x\right) and \text{cos}\left(3x\right)\right\} are orthogonal over the interval \ds \left[ - \pi ,\pi \right].

Answer

\ds \int\limits_{ - \pi }^{\pi }\text{sin}\left(2x\right)\text{cos}\left(3x\right)\,dx =0

50. Let m,n be real numbers such that |m|\ne|n|. Evaluate \ds \int\limits_{ - \pi }^{\pi }\text{sin}\left(mx\right)\text{cos}\left(nx\right)\,dx .

For each pair of integrals, determine which one is easier to evaluate. Explain your reasoning.

51. \ds \int {\text{tan}}^{350}(x)\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx or \ds \int {\text{tan}}^{350}(x)\phantom{\rule{0.1em}{0ex}}\text{sec}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Answer

The first integral is easier to evaluate as it can be done just using a substitution evaluated u=\tan(x), while the second integral is of a complicated type when the power of tangent is even and the power of secant is odd.

52. \ds \int {\text{sin}}^{456}(x)\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx or \ds \int {\text{sin}}^{2}(x)\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{2}(x)\phantom{\rule{0.2em}{0ex}}\,dx

Glossary

power reduction formula
a rule that allows an integral of a power of a trigonometric function to be exchanged for an integral involving a lower power
trigonometric integral
an integral involving powers and products of trigonometric functions

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Calculus: Volume 2 (Second University of Manitoba Edition) by Gilbert Strang and Edward 'Jed' Herman, modified by Varvara Shepelska is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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