3.1 Integration by Parts

Learning Objectives

  • Recognize when to use integration by parts.
  • Use the integration-by-parts formula to evaluate indefinite integrals.
  • Use the integration-by-parts formula for definite integrals.

By now we have a fairly thorough procedure for how to evaluate many basic integrals. However, although we can integrate \ds \int x\,\phantom{\rule{0.1em}{0ex}}\text{sin}\left({x}^{2}\right)\,dx by using the substitution, \ds u={x}^{2}, something as simple looking as \ds \int x\phantom{\rule{0.1em}{0ex}}\,\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx defies us. Although there is no product rule for integration, there is a technique based on the product rule for differentiation that allows us to exchange one integral for another. We call this technique integration by parts.

The Integration-by-Parts Formula

If \ds h\left(x\right)=f\left(x\right)g\left(x\right), then by using the product rule, we obtain \ds {h}^{\prime }\left(x\right)={f}^{\prime }\left(x\right)g\left(x\right)+{g}^{\prime }\left(x\right)f\left(x\right).

We rewrite this formula as \ds f(x)g'(x)=h'(x)-g(x)f'(x) and then integrate both sides of the last equation:

\ds \int f(x)g'(x)\,dx =\int \left(h'(x)-g(x)f'(x)\right)\,dx =\int h'(x)\,dx-\int g(x)f'(x)\,dx .

Since h(x)=f(x)g(x) is an antiderivative of h'(x), we have

\ds \int f\left(x\right){g}^{\prime }\left(x\right)\,dx =f\left(x\right)g\left(x\right)-\int g\left(x\right){f}^{\prime }\left(x\right)\,dx .

By making the substitutions \ds u=f\left(x\right) and \ds v=g\left(x\right), which in turn make \ds du={f}^{\prime }\left(x\right)\,dx and \ds dv={g}^{\prime }\left(x\right)\,dx , we have the more compact form

\ds \int u\phantom{\rule{0.2em}{0ex}}dv=uv-\int v\phantom{\rule{0.2em}{0ex}}du.

Integration by Parts

Let \ds u=f\left(x\right) and \ds v=g\left(x\right) be functions with continuous derivatives. Then, the integration-by-parts formula for the integral involving these two functions is:

\ds (*)\quad \int u\phantom{\rule{0.2em}{0ex}}dv=uv-\int v\phantom{\rule{0.2em}{0ex}}du.

The advantage of using the integration-by-parts formula is that we can exchange one integral for another, possibly easier, integral. The following example illustrates its use.

Using Integration by Parts

Use integration by parts with \ds u=x and \ds dv=\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx to evaluate \ds \int x\phantom{\rule{0.1em}{0ex}}\,\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx .


By choosing \ds u=x, we have \ds du=1\,dx . Since \ds dv=\text{sin}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\,dx , we can take v to be any antiderivative of \sin(x), and the simplest choice would be \ds v=- \text{cos}\phantom{\rule{0.1em}{0ex}}(x). It is handy to keep track of these values as follows:

\ds \begin{array}{ccccccc}\hfill u&\ds =\hfill &\ds x\hfill &\ds &\ds \hfill dv&\ds =\hfill &\ds \text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill \\[3mm]\ds \hfill du&\ds =\hfill &\ds 1\,dx \hfill &\ds &\ds \hfill v&\ds =\hfill &\ds - \text{cos}\phantom{\rule{0.1em}{0ex}}(x).\hfill \end{array}

Applying the integration-by-parts formula (*) results in

\ds \begin{array}{cccc}\hfill \ds\int x\,\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx &\ds =\left(x\right)\left( - \text{cos}\phantom{\rule{0.1em}{0ex}}(x)\right)-\int \left( - \text{cos}\phantom{\rule{0.1em}{0ex}}(x)\right)\left(1\,dx \right)\hfill &\ds &\ds \text{substitute}\hfill \\[5mm]\ds &\ds = - x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)+\int \text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds &\ds \text{simplify}\hfill \\[5mm]\ds &\ds = - x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)+\text{sin}\phantom{\rule{0.1em}{0ex}}(x)+C.\hfill &\ds &\ds \text{integrate} \cos(x)\hfill \end{array}


At this point, there are probably a few items that need clarification. First of all, you may be curious about what would have happened if we had chosen \ds u=\text{sin}\phantom{\rule{0.1em}{0ex}}(x) and \ds dv=x. If we had done so, then we would have \ds du=\text{cos}\phantom{\rule{0.1em}{0ex}}(x) and \ds v=\frac{1}{2}{x}^{2}. Thus, after applying integration by parts, we would get \ds {\int }^{\text{​}}x\phantom{\rule{0.1em}{0ex}}\,\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx =\frac{1}{2}{x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)-{\int }^{\text{​}}\frac{1}{2}{x}^{2}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx . Unfortunately, with the new integral, we are in no better position than before. It is important to keep in mind that when we apply integration by parts, we may need to try several choices for \ds u and \ds dv before finding a choice that works.

Second, you may wonder why, when we find \ds v as an antiderivative of \sin(x) we do not use \ds v= - \text{cos}\phantom{\rule{0.1em}{0ex}}(x)+K. To see that it makes no difference, we can rework the problem using \ds v= - \text{cos}\phantom{\rule{0.1em}{0ex}}(x)+K\text{:}

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

As you can see, it makes no difference in the final solution.

Last, we can verify that our antiderivative is correct by differentiating \ds - x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)+\text{sin}\phantom{\rule{0.1em}{0ex}}(x)+C\text{:}

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

Therefore, the answer we obtained is correct.

Evaluate \ds {\int }^{\text{​}}x{e}^{2x}\,dx using the integration-by-parts formula with \ds u=x and \ds dv={e}^{2x}\,dx .


\ds {\int }^{\text{​}}x{e}^{2x}\,dx =\frac{1}{2}x{e}^{2x}-\frac{1}{4}{e}^{2x}+C


Find \ds du and \ds v, and use the previous example as a guide.

The natural question to ask at this point is: How do we know how to choose \ds u and \ds dv? Sometimes it is a matter of trial and error; however, the acronym LIATE can often help to take some of the guesswork out of our choices. This acronym stands for Logarithmic Functions, Inverse Trigonometric Functions, Algebraic Functions, Trigonometric Functions, and Exponential Functions. This mnemonic serves as an aid in determining an appropriate choice for \ds u.

The type of function in the integral that appears first in the list should be our first choice of \ds u. For example, if an integral contains a logarithmic function and an algebraic function, we should choose \ds u to be the logarithmic function, because L comes before A in LIATE. The integral in \ds\int x\sin(x)\,dx has a trigonometric function \ds \text{sin}\phantom{\rule{0.1em}{0ex}}(x) and an algebraic function \ds x. Because A comes before T in LIATE, we chose \ds u to be the algebraic function. When we have chosen \ds u, \ds dv is selected to be the remaining part of the function to be integrated, together with \ds \,dx .

Why does this mnemonic work? Remember that whatever we pick to be \ds dv must be something we can integrate. Since we do not have integration formulas that allow us to integrate simple logarithmic functions and inverse trigonometric functions, it makes sense that they should not be chosen as values for \ds dv. Consequently, they should be at the head of the list as choices for \ds u. Thus, we put LI at the beginning of the mnemonic. (We could just as easily have started with IL, since these two types of functions won’t appear together in an integration-by-parts problem.) The exponential and trigonometric functions are at the end of our list because they are fairly easy to integrate and make good choices for \ds dv. Thus, we have TE at the end of our mnemonic. (We could just as easily have used ET at the end, since when these types of functions appear together it usually doesn’t really matter which one is \ds u and which one is \ds dv.) Algebraic functions are generally easy both to integrate and to differentiate, and they come in the middle of the mnemonic.

Using Integration by Parts

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


Begin by rewriting the integral:

\ds \int \frac{\text{ln}\phantom{\rule{0.1em}{0ex}}(x)}{{x}^{3}}\,dx ={\int }^{\text{​}}{x}^{-3}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

Since this integral contains the algebraic function \ds {x}^{-3} and the logarithmic function \ds \text{ln}\phantom{\rule{0.1em}{0ex}}(x), choose \ds u=\text{ln}\phantom{\rule{0.1em}{0ex}}(x), since L comes before A in LIATE. After we have chosen \ds u=\text{ln}\phantom{\rule{0.1em}{0ex}}(x), we must choose \ds dv={x}^{-3}\,dx .

Next, since \ds u=\text{ln}\phantom{\rule{0.1em}{0ex}}(x), we have \ds du=\frac{1}{x}\,dx . Also, \ds {\int }^{\text{​}}{x}^{-3}\,dx =-\frac{1}{2}{x}^{-2}+C, and so we take \ds v=-\frac12 x^{-2}. Summarizing,

\ds \begin{array}{ccccccc}\hfill u&\ds =\hfill &\ds \text{ln}\phantom{\rule{0.1em}{0ex}}(x)\hfill &\ds &\ds \hfill dv&\ds =\hfill &\ds {x}^{-3}\,dx \hfill \\[3mm]\ds \hfill du&\ds =\hfill &\ds \frac{1}{x}\,dx \hfill &\ds &\ds \hfill v&\ds =\hfill &\ds -\frac{1}{2}{x}^{-2}.\hfill \end{array}

Substituting into (*) gives

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

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


\ds \frac{1}{2}{x}^{2}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)-\frac{1}{4}{x}^{2}+C


Use \ds u=\text{ln}\phantom{\rule{0.1em}{0ex}}(x) and \ds dv=x\phantom{\rule{0.2em}{0ex}}\,dx .

In some cases, as in the next two examples, it may be necessary to apply integration by parts more than once.

Applying Integration by Parts More Than Once

Evaluate \ds {\int }^{\text{​}}{x}^{2}{e}^{3x}\,dx .


Using LIATE, choose \ds u={x}^{2} and \ds dv={e}^{3x}\,dx . Thus, \ds du=2x\phantom{\rule{0.2em}{0ex}}\,dx and \ds \int {e}^{3x}\,dx =\left(\frac{1}{3}\right){e}^{3x}+C, which means we can take \ds v=\frac13e^{3x}. Therefore,

\ds \begin{array}{ccccccc}\hfill u&\ds =\hfill &\ds {x}^{2}\hfill &\ds &\ds \hfill dv&\ds =\hfill &\ds {e}^{3x}\,dx \hfill \\[3mm]\ds \hfill du&\ds =\hfill &\ds 2x\phantom{\rule{0.2em}{0ex}}\,dx \hfill &\ds &\ds \hfill v&\ds =\hfill &\ds \frac{1}{3}{e}^{3x}.\hfill \end{array}

Substituting into (*) produces

\ds \int {x}^{2}{e}^{3x}\,dx =\frac{1}{3}{x}^{2}{e}^{3x}-\int \frac{2}{3}x{e}^{3x}\,dx .

We still cannot integrate \ds \int \frac{2}{3}x{e}^{3x}\,dx directly, but the integral now has a lower power on \ds x. We can evaluate this new integral by using integration by parts again. To do this, choose \ds u=x and \ds dv=\frac{2}{3}{e}^{3x}\,dx . Thus, \ds du=\,dx and \ds \int \left(\frac{2}{3}\right){e}^{3x}\,dx =\left(\frac{2}{9}\right){e}^{3x}, which means we can take \ds v=\frac29 e^{3x}. Now we have

\ds \begin{array}{ccccccc}\hfill u&\ds =\hfill &\ds x\hfill &\ds &\ds \hfill dv&\ds =\hfill &\ds \frac{2}{3}{e}^{3x}\,dx \hfill \\[3mm]\ds \hfill du&\ds =\hfill &\ds \,dx \hfill &\ds &\ds \hfill v&\ds =\hfill &\ds \frac{2}{9}{e}^{3x}.\hfill \end{array}

Going back to the previous equation and using (*) yields

\ds\begin{array}{ll} \ds{\int }^{\text{​}}{x}^{2}{e}^{3x}\,dx &=\ds\frac{1}{3}{x}^{2}{e}^{3x}-\int \frac{2}{3}x{e}^{3x}\,dx \hfill\\[5mm]&=\ds\frac{1}{3}{x}^{2}{e}^{3x}-\left(\frac{2}{9}x{e}^{3x}-{\int }^{\text{​}}\frac{2}{9}{e}^{3x}\,dx \right).\hfill\end{array}

After evaluating the last integral and simplifying, we obtain

\ds \int {x}^{2}{e}^{3x}\,dx =\frac{1}{3}{x}^{2}{e}^{3x}-\frac{2}{9}x{e}^{3x}+\frac{2}{27}{e}^{3x}+C.

Applying Integration by Parts When LIATE Doesn’t Quite Work

Evaluate \ds {\int }^{\text{​}}{t}^{3}{e}^{{t}^{2}}dt.


If we use a strict interpretation of the mnemonic LIATE to make our choice of \ds u, we end up with \ds u={t}^{3} and \ds dv={e}^{{t}^{2}}dt. Unfortunately, this choice won’t work because we are unable to evaluate \ds {\int }^{\text{​}}{e}^{{t}^{2}}dt. However, since we can evaluate \ds {\int }^{\text{​}}t{e}^{{t}^{2}}\,dt, we can try choosing \ds u={t}^{2} and \ds dv=t{e}^{{t}^{2}}dt. We then have

\begin{array}{ll}\ds\int t e^{t^2}\,dt&\ds=\int e^{t^2}\frac12 (2t)\,dt=\frac12\int e^{t^2}(t^2)'\, dt\hfill\\[5mm]&\ds=\frac12\int e^w dw=\frac12e^w+C=\frac12e^{t^2}+C,\end{array}
and so we can take \ds v=\frac12 e^{t^2}.

With these choices we have

\ds \begin{array}{ccccccc}\hfill u&\ds =\hfill &\ds {t}^{2}\hfill &\ds &\ds \hfill dv&\ds =\hfill &\ds t{e}^{{t}^{2}}dt\hfill \\[3mm]\ds \hfill du&\ds =\hfill &\ds 2t\phantom{\rule{0.2em}{0ex}}dt\hfill &\ds &\ds \hfill v&\ds =\hfill &\ds \frac{1}{2}{e}^{{t}^{2}}.\hfill \end{array}

Thus, we obtain

\ds \begin{array}{cc}\ds {\int }^{\phantom{\rule{0.2em}{0ex}}}{t}^{3}{e}^{{t}^{2}}dt\hfill &\ds ={t}^{2}\cdot\frac{1}{2}e^{{t}^{2}}-\int \frac{1}{2}{e}^{{t}^{2}}2tdt\hfill \\[5mm]\ds \hfill &\ds =\frac{1}{2}{t}^{2}{e}^{{t}^{2}}-\frac{1}{2}{e}^{{t}^{2}}+C.\hfill \end{array}

Applying Integration by Parts More Than Once

Evaluate \ds {\int }^{\text{​}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x\right)\,dx .


This integral appears to have only one function—namely, \ds \text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right) —however, we can always use the constant function 1 as the other function. In this example, let’s choose \ds u=\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right) and \ds dv=1\,dx . (The decision to use \ds u=\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right) is easy. We can’t choose \ds dv=\phantom{\rule{0.2em}{0ex}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\,dx because if we could integrate it, we wouldn’t be using integration by parts in the first place!) Consequently, \ds du=\text{cos}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\left(\frac1x\right)\,dx and we can take\ds v=x as an antiderivative of 1. After applying integration by parts to the integral and simplifying, we obtain

\ds {\int }^{\text{​}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\,dx =x\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)-{\int }^{\text{​}}\text{cos}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\,dx .

Unfortunately, this process leaves us with a new integral that is very similar to the original. However, let’s see what happens when we apply integration by parts again. This time let’s choose \ds u=\text{cos}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right) and \ds dv=1\,dx , making \ds du= - \text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\left(\frac1x\right)\,dx and, again, \ds v=x. Substituting, we have

\ds {\int }^{\text{​}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\,dx =x\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)-\left(x\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)-{\int }^{\text{​}}-\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\,dx \right).

After simplifying, we obtain

\ds {\int }^{\text{​}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\,dx =x\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)-x\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)-{\int }^{\text{​}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\,dx .

The last integral is now the same as the original. It may seem that we have simply gone in a circle, but now we can actually evaluate the integral. To see how to do this more clearly, let I be a particular antiderivative of \ds\sin(\ln(x)). Substituting I instead of \ds {\int }^{\text{​}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\,dx into the above equality, we obtain the following equation for I.

\ds I=x\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)-x\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)-I.

To find I, add \ds I to both sides of the equation:

\ds 2I=x\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x\right)-x\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x\right),

and then divide by 2

\ds I=\frac{1}{2}x\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x\right)-\frac{1}{2}x\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x\right).

Since I was a particular  antiderivative of \sin(\ln(x)), we have

\ds {\int }^{\text{​}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\,dx =I+C=\frac{1}{2}x\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)-\frac{1}{2}x\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)+C,

where C is an arbitrary constant.


If this method feels a little strange at first, we can check the answer by differentiation:

\ds \begin{array}{c}\ds\frac{d}{dx }\left(\frac{1}{2}x\phantom{\rule{0.1em}{0ex}}\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)-\frac{1}{2}x\phantom{\rule{0.1em}{0ex}}\text{cos}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\right)=\hfill\\[5mm]\ds =\frac{1}{2}\left(\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\right)+\text{cos}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\cdot \frac{1}{x}\cdot \frac{1}{2}x-\left(\frac{1}{2}\text{cos}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)-\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)\cdot \frac{1}{x}\cdot \frac{1}{2}x\right)\hfill \\[5mm]\ds =\text{sin}\left(\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right).\hfill \end{array}

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


\ds - {x}^{2}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)+2x\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)+2\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)+C


This is similar to \ds\int x^2 e^{3x}\,dx from one of the previous examples.

Integration by Parts for Definite Integrals

Now that we have used integration by parts successfully to evaluate indefinite integrals, we turn our attention to definite integrals. The integration technique is really the same, only we add a step to evaluate the integral at the upper and lower limits of integration.

Integration by Parts for Definite Integrals

Let \ds f\left(x\right) and \ds g\left(x\right) be functions with continuous derivatives on \ds \left[a,b\right]. Then

\ds \int\limits_a^b f\left(x\right){g}^{\prime }\left(x\right)\,dx =f\left(x\right)g\left(x\right)\Big|_a^b-\int\limits_a^b g\left(x\right){f}^{\prime }\left(x\right)\,dx .
If we denote u=f(x) and v=g(x), then it becomes
\ds (**)\quad\int\limits_{a}^{b}u\phantom{\rule{0.2em}{0ex}}dv={uv}\Big|_{a}^{b}-\int\limits_{a}^{b}v\phantom{\rule{0.2em}{0ex}}du,
where the bounds of integration and substitution are specified for the variable x.

Finding the Area of a Region

Find the area of the region bounded above by the graph of \ds y={\text{tan}}^{-1}(x) and below by the x-axis over the interval \ds \left[0,1\right].


This region is shown in Figure 1 below. To find the area, we must evaluate \ds \underset{0}{\overset{1}{\int}}{\text{tan}}^{-1}(x)\phantom{\rule{0.2em}{0ex}}\,dx .

This figure is the graph of the inverse tangent function. It is an increasing function that passes through the origin. In the first quadrant there is a shaded region under the graph, above the x-axis. The shaded area is bounded to the right at x = 1.
Figure 1. To find the area of the shaded region, we have to use integration by parts.

For this integral, let’s choose \ds u={\text{tan}}^{-1}(x) and \ds dv=\,dx , thereby making \ds du=\frac{1}{{x}^{2}+1}\,dx and \ds v=x. After applying the integration-by-parts formula (**) for definite integrals, we obtain

\ds \text{Area}=x\phantom{\rule{0.1em}{0ex}}{\text{tan}}^{-1}{(x)}\Big|_{0}^{1}-\int\limits_0^1\frac{x}{{x}^{2}+1}\,dx .

We use a substitution of w=1+x^2 to evaluate \ds \int\limits_0^1\frac{x}{{x}^{2}+1}\,dx. We have that dw=2x\,dx, and hence x\,dx=\frac12 dw. Also, when x=0, w=1, and when x=1, w=2. It follows that

\ds \int\limits_0^1\frac{x}{{x}^{2}+1}\,dx =\int\limits_1^2 \frac{1}{2}\frac1w\,dw=\frac12\ln|w|\Big|_{1}^{2}.=\frac12\ln(2)


\ds \text{Area}=x\phantom{\rule{0.2em}{0ex}}{\mathrm{tan}}^{-1}(x){}\Big|_{0}^{1}-\int\limits_0^1\frac{x}{{x}^{2}+1}\,dx=\frac{\pi }{4}-\frac{1}{2}\mathrm{ln}\phantom{\rule{0.2em}{0ex}}(2).

Finding a Volume of Revolution

Find the volume of the solid obtained by revolving the region bounded by the graph of \ds f\left(x\right)={e}^{ - x}, the x-axis, the y-axis, and the line \ds x=1 about the y-axis.


We use cylindrical shells method to solve this problem. Begin by sketching the region to be revolved, along with a typical rectangle (see the following graph).

This figure is the graph of the function e^-x. It is an increasing function on the left side of the y-axis and decreasing on the right side of the y-axis. The curve also comes to a point on the y-axis at y=1. Under the curve there is a shaded rectangle in the first quadrant. There is also a cylinder under the graph, formed by revolving the rectangle around the y-axis.
Figure 2. We can use cylindrical shells to find the volume of revolution.

According to the formula, we must evaluate \ds \int\limits_{0}^{1}2\pi x{e}^{ - x}\,dx =2\pi\int\limits_{0}^{1} x{e}^{ - x}\,dx.

To do this, let \ds u=x and \ds dv={e}^{ - x}\,dx. These choices lead to \ds du=\,dx and \ds v=- {e}^{ - x} as an antiderivative of e^{-x}. Using integration by parts, we obtain

\ds \begin{array}{cccc}\hfill \text{Volume}&\ds =2\pi \underset{0}{\overset{1}{\int }}x{e}^{ - x}\,dx =2\pi \left( - x{e}^{ - x}{}\Big|_{0}^{1}+\underset{0}{\overset{1}{\int }}{e}^{ - x}\,dx \right)\hfill &\ds &\ds \hfill \\[5mm]\ds &\ds =-2\pi x{e}^{ - x}{}\Big|_{0}^{1}-2\pi {e}^{ - x}{}\Big|_{0}^{1}\hfill &\ds &\ds \hfill \\[5mm]\ds &\ds =2\pi -\frac{4\pi }{e}.\hfill &\ds &\ds \hfill \end{array}

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


\ds \frac{\pi }{2}-1


Use integration by parts with \ds u=x and \ds dv=\text{cos}\phantom{\rule{0.1em}{0ex}}x\phantom{\rule{0.2em}{0ex}}\,dx .

Key Concepts

  • The integration-by-parts formula allows the exchange of one integral for another, possibly easier, integral.
  • Integration by parts applies to both definite and indefinite integrals.

Key Equations

  • Integration by parts formula
    \ds \int u\phantom{\rule{0.2em}{0ex}}dv=uv-\int v\phantom{\rule{0.2em}{0ex}}du
  • Integration by parts for definite integrals
    \ds \int\limits_{a}^{b}u\phantom{\rule{0.2em}{0ex}}dv={uv}\Big|_{a}^{b}-\int\limits_{a}^{b}v\phantom{\rule{0.2em}{0ex}}du


In using the technique of integration by parts, you must carefully choose which expression is u. For each of the following exercises, use the guidelines in this section to choose u. Do not evaluate the integrals.

1. \ds \int {x}^{3}{e}^{2x}\,dx


\ds u={x}^{3}

2. \ds \int \sqrt x\,\text{ln}\left(x\right)\,dx

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


\ds u={y}^{5}

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

5. \ds \int {e}^{3x}\text{sin}\left(2x\right)\,dx


\ds u=\text{sin}\left(2x\right) or u=e^{3x}; this example requires two applications of integration by parts, and it is important to choose u in the same way in both of them.

For the following exercises, evaluate the integral using integration by parts.

6. \ds \int v\phantom{\rule{0.1em}{0ex}}\text{sin}\phantom{\rule{0.1em}{0ex}}(v)\,dv

7. \ds \int \text{ln}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

(Hint: \ds \int \text{ln}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx is equivalent to \ds \int 1\cdot \text{ln}\left(x\right)\,dx .\right))


\ds - x+x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)+C

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

9. \ds \int {\text{cos}}^{-1}(x)\phantom{\rule{0.2em}{0ex}}\,dx


\ds x\phantom{\rule{0.1em}{0ex}}{\text{cos}}^{-1}(x)-\sqrt{1-x^2}+C

10. \ds \int ({x}^{2}-2x){e}^{x}\,dx

11. \ds \int x\phantom{\rule{0.1em}{0ex}}\text{sin}\left(2x\right)\,dx


\ds -\frac{1}{2}x\phantom{\rule{0.1em}{0ex}}\text{cos}\left(2x\right)+\frac{1}{4}\text{sin}\left(2x\right)+C

12. \ds \int x{e}^{4x}\,dx

13. \ds \int (3x-2){e}^{ - x}\,dx


\ds {e}^{ - x}\left(-1-3x\right)+C

14. \ds \int \sqrt{x^5}\ln(x)\,dx

15. \ds \int (1-{x}^{2})\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx


\ds (3-x^2)\sin(x)-2x\cos(x)+C

16. \ds \int \frac{\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}(x^2)}{x^2}\phantom{\rule{0.2em}{0ex}}\,dx

17. \ds \int \text{ln}\left(2x+1\right)\,dx


\ds \frac{1}{2}\left(1+2x\right)\left(-1+\text{ln}\left(1+2x\right)\right)+C

18. \ds \int \frac{1-x}{{e}^{2x}}\,dx

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


\ds \frac{1}{2}{e}^{x}\left( - \text{cos}\phantom{\rule{0.1em}{0ex}}x+\text{sin}\phantom{\rule{0.1em}{0ex}}x\right)+C

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

21. \ds \int \text{sin}\left(\text{ln}\left(2x\right)\right)\,dx


\ds -\frac{1}{2}x\phantom{\rule{0.1em}{0ex}}\text{cos}\left[\text{ln}\left(2x\right)\right]+\frac{1}{2}x\phantom{\rule{0.1em}{0ex}}\text{sin}\left[\text{ln}\left(2x\right)\right]+C

22. \ds \int \cos\left(\text{ln}\phantom{\rule{0.1em}{0ex}}x\right)\,dx

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


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

24. \ds \int \text{ln}\left(\sqrt x\right)\,dx

(Hint: you can simplify the integrand using properties of logarithms.)

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


\ds \left( - \phantom{\rule{0.2em}{0ex}}\frac{{x}^{3}}{9}+\frac{1}{3}{x}^{3}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\right)+C

26. \ds \int {x\text{sin}}^{-1}\left(\frac1x\right)\phantom{\rule{0.2em}{0ex}}\,dx

27. \ds \int {\text{cos}}^{-1}\left(2x\right)\,dx


\ds -\frac{1}{2}\sqrt{1-4{x}^{2}}+x\phantom{\rule{0.2em}{0ex}}{\text{cos}}^{-1}\left(2x\right)+C

28. \ds \int x\phantom{\rule{0.1em}{0ex}}\text{arctan}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

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


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

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

31. \ds \int x\phantom{\rule{0.1em}{0ex}}{\text{sec}}^{-1}(x)\phantom{\rule{0.2em}{0ex}}\,dx


\ds \frac{1}{2}x\left( - \sqrt{1-\frac{1}{{x}^{2}}}+x\cdot {\text{sec}}^{-1}(x)\right)+C

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

33. \ds \int 2x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}(x-1)\phantom{\rule{0.2em}{0ex}}\,dx


\ds (x^2-1)\ln(x-1)-\frac12 x^2-x+C

For the following exercises, compute the given definite integral using integration by parts.

34. \ds \int\limits_{1\text{/}e}^{1}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

35. \ds \int\limits_{0}^{1}x{e}^{-2x}\,dx


\ds \frac{1}{4}-\frac{3}{4{\text{e}}^{2}}

36. \ds \int\limits_{0}^{1}{e}^{\sqrt{x}}\,dx

(Hint: make a substitution u=\sqrt{x})

37. \ds \int\limits_{1}^{e}(2x-1)\text{ln}\left({x}^{2}\right)\,dx



38. \ds \int\limits_{-\pi}^{0 }x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

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


\ds \pi^2-\pi-2

40. \ds \int\limits_{-1}^{0}\text{ln}\left(1-x\right)\,dx

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


\ds \frac{e^{\pi\text{/}2}+1}2

42. \ds \int\limits_{0}^{1}x\,{5}^{x}\,dx

43. Evaluate \ds \int\limits_0^{1\text{/}2} \sin^{-1}(x)\,dx


\ds \frac{\pi}{12}+\frac{\sqrt3}2-1

Assume that n is a positive integer. Derive the following formulas using integration by parts. These formulas are called reduction formulas because the exponent in the x term has been reduced by one in each case.

44. \ds \int {x}^{n}{e}^{x}\,dx ={x}^{n}{e}^{x}-n\int {x}^{n-1}{e}^{x}\,dx

45. \ds \int {x}^{n}\text{cos}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx ={x}^{n}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)-n\int {x}^{n-1}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx


Take u=x^n and dv=\cos(x)\,dx.

46. \ds \int {x}^{n}\text{sin}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx =\underset{}{______}

47. Integrate \ds \int 2x\sqrt{2x-3}\,dx using two methods:

  1. Using parts, letting \ds dv=\sqrt{2x-3}\phantom{\rule{0.1em}{0ex}}\,dx
  2. Substitution, letting \ds u=2x-3


\ds \frac{2}{5}\left(1+x\right){\left(-3+2x\right)}^{3\text{/}2}+C

State whether you would use integration by parts to evaluate the integral. If so, identify u and dv. If not, describe the technique you would use. You don’t have to evaluate the integral.

48. \ds \int x\phantom{\rule{0.1em}{0ex}}\text{ln}\phantom{\rule{0.1em}{0ex}}(x)\phantom{\rule{0.2em}{0ex}}\,dx

49. \ds \int \frac{{\text{ln}}^{2}(x)}{x}\,dx


No need to use integration by parts. Make a substituion \ds u= \text{ln}\phantom{\rule{0.1em}{0ex}}(x).

50. \ds \int x{e}^{x}\,dx

51. \ds \int x{e}^{{x}^{2}-3}\,dx


No need to use integration by parts. Make a substitution \ds u={x}^{2}-3.

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

53. \ds \int {x}^{2}\text{sin}\left(3{x}^{3}+2\right)\,dx


No need to use integration by parts. Make a substitution \ds u=3{x}^{3}+2.

For the following exercises, sketch the region bounded by the given curves and find its area using integration by parts.

54. \ds y=2x{e}^{ - x},\ y=0,\ x=1

55*. \ds y={\text{sec}}^{3}(x),\ y=0,\ x=0,\ x=\frac{\pi}3.

(Hint: take u=\sec(x) and dv=\sec^2(x)\,dx.)



56. \ds y=\left(x-2\right){e}^{x},\ y=0,\ x=2,\ x=5.

57. \ds y=x\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x),\ y=0,\ x=\frac{11\pi }{2},\  x=\frac{13\pi }{2}.


\ds 12\pi

58. \ds y={e}^{ - x}\text{sin}\left(\pi x\right)

For the following exercises, find the volume generated by rotating the region bounded by the given curves about the specified axes.

59. \ds y=4\phantom{\rule{0.1em}{0ex}}\text{cos}\phantom{\rule{0.1em}{0ex}}(x),\ y=0,\ x=\frac{\pi }{2},\ x=\frac{3\pi }{2}, about the x-axis.


\ds 8{\pi }^{2}

60. \ds y=\text{sin}\phantom{\rule{0.1em}{0ex}}(x),y=0,x=2\pi ,\ x=3\pi about the y-axis

61. \ds y={e}^{ - x}, \ds y=0,\ x=-1,\ x=0 about \ds x=1


\ds 2\pi e

62. \ds y=\text{ln}\phantom{\rule{0.1em}{0ex}}(x),\ y=0,\ x={e}^{2} about the x-axis.

63. \ds y={e}^{x},\ y=0,\ x=0,\ x=\text{ln}\left(7\right) about the y-axis.


\ds 2\pi (7\ln(7)-6)

64. A particle moving along a straight line has a velocity of \ds v\left(t\right)={t}^{2}{e}^{ - t} after t sec. How far does it travel in the first 2 sec?


integration by parts
a technique of integration that allows the exchange of one integral for another using the formula \ds {\int }^{\text{​}}u\phantom{\rule{0.2em}{0ex}}dv=uv-{\int }^{\text{​}}v\phantom{\rule{0.2em}{0ex}}du


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Calculus: Volume 2 (Second University of Manitoba Edition) Copyright © 2021 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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