A powerful technique of integration is integration by parts, which hangs on the formula for differential of a product. Let $$u$$ and $$v$$ be two functions of $$x$$. Then $d(uv)=udv+vdu.$ Taking $$v\,du$$ to the other side, we get $u\,dv=d(uv)-v\,du.$ Integrating both sides, we have $\int udv=\int d(uv)+\int vdu.$ Since $$\int d(uv)=uv+C$$, we get  $\boxed{\int udv=uv-\int vdu.}\tag{a}$Because we have to add a constant when we integrate $$\int vdu$$, we can ignore the constant of integration $$C$$ when we integrate $$\int d(uv)$$.

When we choose expressions for $$u$$ and $$v$$, we usually want $$dv$$ to be an expression that can be easily integrated, $$u$$ to be a function that has a simpler derivative, and $$\int vdu$$ to be easier than $$\int vdu$$.

1. Identify $$u$$ and $$dv$$. It is often successful if we choose for $$u$$ the function in the integrand whose type appears earlier in the LIATE list.1 LIATE is an acronym that stands for

1. Logarithmic functions (like $$\ln x$$)

2. Inverse trigonometric functions (like $$\arcsin x$$)

3. Algebraic functions (like $$3x^{2}+7$$)

4. Trigonometric functions (like $$\sin x$$)

5. Exponential functions (like $$e^{ax}$$)

After choosing $$u$$, whatever left is $$dv$$.

2. Compute $$du$$ and $$v$$.

3. Apply formula (a) to evaluate the integral.

• Note that $$dx$$ is always part of $$dv$$.

#### Show why

Again, generally speaking, we choose $$u$$ and $$dv$$ so that $$u$$ becomes simpler when differentiated and $$dv$$ can be easily integrated. The derivatives of the logarithmic and inverse trigonometric functions are algebraic functions, which are much simpler to deal with. Also, we don’t know how to integrate them directly, so they cannot be chosen as a part of $$dv$$. Therefore, the logarithmic and inverse trigonometric functions are great choices for $$u$$. For example, the derivative of $$\ln x$$ is simply $$1/x$$ and the derivative of $$\arctan x$$ is $$1/(1+x^{2})$$, but how can we integrate $$\ln x$$ or $$\arctan x$$? In the LIATE list, next, we have algebraic functions. The derivative of an algebraic function is another algebraic function, and the derivative is often simpler. For example, the derivative of $$x$$ is simply $$1$$. But the integral an algebraic function sometimes is non-algebraic. For example, the integral of $$1/(1+x^{2})$$ is $$\arctan x+C$$. So we would probably rather differentiate them than integrate them. The last choices for $$u$$ are the trigonometric and exponential functions because the derivative of a trigonometric function is another trigonometric function, and the derivative of an exponential function is still exponential.

Example 1

Evaluate $${\displaystyle \int\ln x\,dx}.$$

Solution 1

Obviously we cannot put $$u=1$$ and $$dv=\ln x\,dx$$ because we do not know how to find $$v$$ in this case. So we are left with one option: $$u=\ln x$$ and $$dv=dx$$. Also, according to LIATE, we should choose $$u=\ln x$$. Then $u=\ln x\Rightarrow du=\frac{1}{x}dx$ $dv=dx\Rightarrow v=\int dx=x+C_{1}$ Integration by parts gives: \begin{aligned} \int\overbrace{\ln x}^{u}\overbrace{dx}^{dv} & =\overbrace{(x+C_{1})\ln x}^{uv}-\int\overbrace{(x+C_{1})}^{v}\overbrace{\frac{1}{x}dx}^{du}\\ & =x\ln x+C_{1}\ln x-\int dx-\int\frac{C_{1}}{x}dx\\ & =x\ln x+\cancel{C_{1}\ln x}-x-\cancel{C_{1}\ln|x|}+C_{2}\\ & =x\ln x-x+C_{2}.\end{aligned} [Because the integrand, $$\ln x$$, makes sense only if $$x>0$$, on the right-hand side, we have $$\ln|x|=\ln x$$]

• In the above example, we see that the constant of integration $$C_{1}$$ when determining $$v$$ as $$\int dv$$ does not appear in the final result. This is, in general, true, for if we write $$v+C_{1}$$ in place of $$v$$ in formula (a), then we have \begin{aligned} \int udv & =u(v+C_{1})-\int(v+C_{1})du\\ & =uv+C_{1}u-C_{1}u-\int vdu\\ & =uv-\int vdu.\end{aligned} So we do not need the most general $$v$$, and some $$v$$ suffices. However, occasionally taking $$C_{1}$$ equal to some specific constant can simplify the calculations. For instance, see Examples 6 and 7.

Example 2

Evaluate $${\displaystyle \int x\sin x\,dx}$$.

Solution 2

The LIATE strategy suggests that we should let $u=x,\qquad\text{and}\qquad dv=\sin x\,dx.$ Then $du=dx,\qquad\text{and}\qquad v=-\cos x.$ Integration by parts yields \begin{aligned} \int x\sin xdx & =\underbrace{-x\cos x}_{uv}-\int\underbrace{(-\cos x)dx}_{vdu}\\ & =-x\cos x+\sin x+C.\end{aligned}

Example 3

Evaluate $${\displaystyle \int x\ln x\,dx}.$$

Solution 3

Method 1: According to the LIATE strategy, the priority for choosing $$u$$ is logarithmic functions. So we let $u=\ln x,\qquad\qquad v=x\,dx.$ Then $du=\frac{1}{x}dx,\qquad\qquad v=\frac{1}{2}x^{2}.$ Integration by parts gives \begin{aligned} \int x\ln xdx & =\overbrace{\frac{1}{2}x^{2}\ln x}^{uv}-\int\overbrace{\frac{1}{2}x^{2}}^{v}\overbrace{\frac{dx}{x}}^{du}\\ & =\frac{1}{2}x^{2}\ln x-\frac{1}{4}x^{2}+C.\end{aligned} Method 2: Let $u=x\qquad\text{and}\qquad dv=\ln x\,dx.$ Then $u=x\Rightarrow du=dx$ $dv=\ln xdx\Rightarrow v=x\ln x-x\qquad\text{(from Example 1)}$ Integrating by parts, we obtain \begin{aligned} \int x\ln xdx & =x(x\ln x-x)-\int(x\ln x-x)dx\\ & =x^{2}\ln x-x^{2}-\int x\ln xdx+\underbrace{\int xdx}_{\frac{x^{2}}{2}+C}\end{aligned} Here $$\int x\ln xdx$$ appears both on the right and left hand sides. Thus $\Rightarrow2\int x\ln xdx=x^{2}\ln x-x^{2}+\frac{1}{2}x^{2}+C$ Finally $\int x\ln xdx=\frac{1}{2}x^{2}\ln x-\frac{1}{4}x^{2}+C_{1},$ where $$C_{1}=C/2$$. Obviously the first method is easier, as in the second method we need to use integration by parts twice.

Example 4

Evaluate $${\displaystyle \int\arctan x\,dx}$$.

Solution 4

Obviously we cannot put $$u=1$$ and $$dv=\arctan xdx$$ because we do not know how to find $$v$$ in this case. So we have to let $u=\arctan x,\qquad\qquad dv=dx.$ Then $du=\frac{1}{1+x^{2}}dx,\qquad\qquad v=x.$ Therefore, $\int\arctan xdx=\underbrace{x\arctan x}_{uv}-\int\frac{x}{1+x^{2}}dx.$ To evaluate the last integral, we put $$t=1+x^{2}$$. Then $$dt=2xdx$$, and \begin{aligned} \int\frac{x}{1+x^{2}}dx & =\int\frac{1}{t}\frac{dt}{2}\\ & =\frac{1}{2}\ln|t|+C\\ & =\frac{1}{2}\ln(1+x^{2})+C &&\small{(t=1+x^{2}>0)}\\ & =\ln\sqrt{1+x^{2}}+C\end{aligned} Finally $\int\arctan x\,dx=x\arctan x-\ln\sqrt{1+x^{2}}+C_{1},$ where $$C_{1}=-C.$$

Example 5

Evaluate $${\displaystyle \int\arcsin x\,dx}.$$

Solution 5

Similar to the previous example, we let $u=\arcsin x,\qquad\qquad dv=dx.$ Then $du=\frac{1}{\sqrt{1-x^{2}}}dx,\qquad\qquad v=x.$ Therefore, $\int\arcsin x\,dx=\underbrace{\arcsin x\ x}_{uv}-\int\underbrace{x}_{v}\underbrace{\frac{1}{\sqrt{1-x^{2}}}dx}_{du}.$ To evaluate $$\int\frac{x}{\sqrt{1-x^{2}}}dx$$, because $$-\frac{1}{2}\frac{d}{dx}(1-x^{2})=x$$ appears in the numerator, we let $$t=1-x^{2}$$. Then $$dt=-2x$$ and \begin{aligned} \int\frac{x}{\sqrt{1-x^{2}}}dx & =\int\frac{\overbrace{-\frac{1}{2}dt}^{xdx}}{\sqrt{t}}\\ & =\int-\frac{1}{2}t^{-1/2}dt\\ & =\frac{-1}{2}(2t^{1/2})+C\\ & =-\sqrt{1-x^{2}}+C.&&\small{(t=1-x^{2})}\end{aligned} Finally \begin{aligned} \int\arcsin x\,dx & =x\arcsin x-\left(-\sqrt{1-x^{2}}+C\right)\\ & =x\arcsin x+\sqrt{1-x^{2}}+C_{1},\end{aligned} where $$C_{1}=-C$$.

Example 6

Evaluate $${\displaystyle \int x\tan^{-1}x\,dx}$$. [Notice that $$\tan^{-1}x$$ is the same as $$\arctan x$$]

Solution 6

The LIATE strategy suggests that we let $u=\tan^{-1}x,\qquad dv=x\,dx.$ Notice that because it is easy to differentiate $$\tan^{-1}x$$ but hard to integrate, and also because its derivative is algebraic; it is the best choice for $$u$$. So $du=\frac{1}{1+x^{2}}dx,\qquad\qquad v=\frac{1}{2}x^{2}.$ and \begin{aligned} \int x\tan^{-1}x\,dx & =\overbrace{\frac{1}{2}x^{2}\tan^{-1}x}^{uv}-\int\overbrace{\frac{x^{2}}{2(1+x^{2})}dx}^{vdu}\\ & =\frac{1}{2}x^{2}\tan^{-1}x-\frac{1}{2}\int\frac{(1+x^{2})-1}{1+x^{2}}dx\\ & =\frac{1}{2}x^{2}\tan^{-1}x-\frac{1}{2}\int dx+\frac{1}{2}\int\frac{dx}{1+x^{2}}\\ & =\frac{1}{2}x^{2}\tan^{-1}x-\frac{1}{2}x+\frac{1}{2}\tan^{-1}x+C\\ & =\frac{1}{2}(x^{2}+1)\tan^{-1}x-\frac{1}{2}x+C.\end{aligned} The same result is obtained more quickly if we take $v=\frac{1}{2}x^{2}+C_{1}.$ Then integration by parts gives $\int x\tan^{-1}x\,dx=\left(\frac{1}{2}x^{2}+C_{1}\right)\tan^{-1}x-\int\frac{\frac{1}{2}x^{2}+C_{1}}{1+x^{2}}dx.$ Now taking $$C_{1}=1/2$$ simplifies the last integral and leads to the following result: \begin{aligned} \int x\tan^{-1}x & =\frac{1}{2}\left(x^{2}+1\right)\tan^{-1}x-\frac{1}{2}\int\overset{1}{\cancel{\frac{x^{2}+1}{1+x^{2}}}}dx\\ & =\frac{1}{2}\left(x^{2}+1\right)\tan^{-1}x-\frac{1}{2}x+C.\end{aligned}

Example 7

Evaluate $${\displaystyle \int\tan^{-1}\sqrt{x}\:dx}$$.

Solution 7

Let $u=\tan^{-1}\sqrt{x},\qquad\text{and}\qquad dv=dx.$ Then $u=\frac{1}{1+(\sqrt{x})^{2}}\frac{1}{2\sqrt{x}}dx,\qquad\qquad v=x+C_{1}.$ and $\int\tan^{-1}\sqrt{x}\,dx=(x+C_{1})\tan^{-1}\sqrt{x}-\int(x+C_{1})\frac{1}{1+x}\frac{1}{2\sqrt{x}}dx.$ Taking $$C_{1}=1$$ simplifies the second integral. So \begin{aligned} \int\tan^{-1}\sqrt{x}dx & =(x+1)\tan^{-1}\sqrt{x}-\int\frac{1}{2\sqrt{x}}dx\\ & =(x+1)\tan^{-1}\sqrt{x}-\sqrt{x}+C.\end{aligned}

Example 8

Evaluate $${\displaystyle \int\frac{x}{\cos^{2}x}dx}$$.

Solution 8
Because we can easily integrate $$\frac{1}{\cos^{2}x}=\sec^{2}x$$, we let $u=x,\qquad\qquad dv=\frac{1}{\cos^{2}x}dx=\sec^{2}x\ dx.$ Notice that this choice for $$u$$ agrees with the LIATE strategy. So $u=x\quad\Rightarrow\quad du=dx$ $dv=\sec^{2}x\ dx\quad\Rightarrow\quad v=\tan x,$ and $\int\frac{x}{\cos^{2}x}dx=\underbrace{x\tan x}_{uv}-\int\underbrace{\tan x}_{v}\underbrace{dx}_{du}.$ We know $\int\tan xdx=\int\frac{\sin x}{\cos x}dx=-\int\frac{d(\cos x)}{\cos x}=-\ln|\cos x|+C,$ so \begin{aligned} \int\frac{x}{\cos^{2}x}dx & =x\tan x-(-\ln|\cos x|+C)\\ & =x\tan x+\ln|\cos x|+C_{1},\end{aligned} with $$C_{1}=-C.$$

Example 9

Apply integration by parts to evaluate $${\displaystyle \int\sqrt{a^{2}-x^{2}}dx}$$.

Solution 9

Let $u=\sqrt{a^{2}-x^{2}},\qquad\qquad dv=dx.$ Then $du=\frac{1}{2\sqrt{a^{2}-x^{2}}}\cdot\dfrac{d}{dx}(a^{2}-x^{2})=-\frac{x}{\sqrt{a^{2}-x^{2}}},\qquad v=x.$ Integration by parts gives $\int\overbrace{\sqrt{a^{2}-x^{2}}}^{u}\overbrace{dx}^{dv}=\overbrace{x\sqrt{a^{2}-x^{2}}}^{uv}-\int\overbrace{-\frac{x^{2}}{\sqrt{a^{2}-x^{2}}}dx}^{vdu}.$ We may write $\frac{x^{2}}{\sqrt{a^{2}-x^{2}}}=-\frac{a^{2}-x^{2}}{\sqrt{a^{2}-x^{2}}}+\frac{a^{2}}{\sqrt{a^{2}-x^{2}}}=-\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{\sqrt{a^{2}-x^{2}}}.$ We have therefore \begin{aligned} \int\sqrt{a^{2}-x^{2}}dx & =x\sqrt{a^{2}-x^{2}}+\int\left(-\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{\sqrt{a^{2}-x^{2}}}\right)dx\\ & =x\sqrt{a^{2}-x^{2}}-\int\sqrt{a^{2}-x^{2}}dx+a^{2}\int\frac{d\left(\frac{x}{a}\right)}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}}.\end{aligned} Finally $2\int\sqrt{a^{2}-x^{2}}dx=x\sqrt{a^{2}-x^{2}}+a^{2}\int\frac{d\left(\frac{x}{a}\right)}{\sqrt{1-\left(\frac{x}{a}\right)^{2}}}$ and $\int\sqrt{a^{2}-x^{2}}dx=\frac{1}{2}x\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\arcsin\left(\frac{x}{a}\right).$

Example 10

Evaluate $${\displaystyle \int\sqrt{x}\ln x\,dx}$$.

Solution 10

The LIATE strategy suggests that we let $u=\ln x\qquad\text{and}\qquad dv=\sqrt{x}dx.$ So $du=\frac{1}{x}dx\qquad v=\int\sqrt{x}dx=\frac{2}{3}x^{3/2}$ and \begin{aligned} \int\overbrace{\ln x}^{u}\overbrace{\sqrt{x}dx}^{dv} & =\overbrace{\frac{2}{3}x^{3/2}\ln x}^{uv}-\int\overbrace{\frac{2}{3}x^{3/2}}^{v}\overbrace{x^{-1}dx}^{du}\\ & =\frac{2}{3}x^{3/2}\ln x-\frac{2}{3}\int x^{1/2}dx\\ & =\frac{2}{3}x^{3/2}\ln x-\frac{4}{9}x^{3/2}+C.\end{aligned}

• Sometimes, first, we need to use integration by substitution, and then apply integration by parts.

Example 11

Evaluate $${\displaystyle \int\sin\sqrt{x}\:dx}$$.

Solution 11

Let $$\theta=\sqrt{x}$$. Then $d\theta=\frac{1}{2\sqrt{x}}dx\quad\text{or}\quad2\underbrace{\sqrt{x}}_{\theta}d\theta=dx.$ Therefore, \begin{aligned} \int\sin\sqrt{x}\:dx & =\int\sin\theta\underbrace{2\theta\,d\theta}_{dx}\\ & =2\int\theta\sin\theta\,d\theta.\end{aligned} In Example 2, we learned that $$\int x\sin xdx=-x\cos x+\sin x+C$$. Therefore, \begin{aligned} \int\sin\sqrt{x}\,dx & =2(-\theta\cos\theta+\sin\theta)+C_{1}\\ & =-2\sqrt{x}\cos\sqrt{x}+2\sin\sqrt{x}+C_{1}. &&\small{(\theta=\sqrt{x})}\end{aligned}

Example 12

Evaluate $${\displaystyle \int\arctan\sqrt{x}\:dx}$$.

Solution 12

Let $$t=\sqrt{x}$$. Then $dt=\frac{1}{2\sqrt{x}}dx,$ or $dx=2\sqrt{x}dt=2tdt$ Therefore, $\int\arctan\sqrt{x}\,dx=2\int t\arctan t\,dt.$ We had already obtained the second integral in Example [exa:Ch7-Integration-byParts-x*arctan]. So \begin{aligned} \int\arctan\sqrt{x}dx & =2\int t\arctan t\,dt\\ & =2\left[\frac{1}{2}\left(t^{2}+1\right)\arctan t-\frac{1}{2}t\right]+C\\ & =(x+1)\arctan\sqrt{x}-\sqrt{x}+C.\end{aligned}

To evaluate $$\int f(x)dx$$ using integration by parts, the LIATE strategy fails when $$f(x)$$ is an analytic function (because all we have is analytic). However, if we can decompose $$f(x)dx$$ into two factor $$P(x)$$ and $$Q(x)dx$$ such that we can integrate $$Q(x)dx$$, we may set $$u=P(x)$$ and $$dv=Q(x)dx$$ and try integration by parts. The following shows two examples of such situations.

Example 13

Evaluate $${\displaystyle \int x^{3}\sqrt{a^{2}+x^{2}}}dx$$.

Solution 13

This may be divided into two factors $u=x^{2}\qquad dv=x\sqrt{a^{2}+x^{2}}dx,$ of which the second can be integrated: \begin{aligned} \int x\sqrt{a^{2}+x^{2}}dx & =\int\sqrt{z}\overbrace{\frac{1}{2}zdz}^{xdx}&& {\small(z=a^{2}+x^{2})}\\ & =\frac{1}{2}\cdot\frac{2}{3}z^{3/2}+C_{1}\\ & =\frac{1}{3}(a^{2}+x^{2})^{3/2}+C_{1}.\end{aligned} So $u=x^{2}\Rightarrow du=2xdx$ $dv=x\sqrt{a^{2}+x^{2}}dx\Rightarrow v=\frac{1}{3}(a^{2}+x^{2})^{\frac{3}{2}}$ and $\int x^{3}\sqrt{a^{2}+x^{2}}dx=\overbrace{\frac{1}{3}x^{2}(a^{2}+x^{2})^{\frac{3}{2}}}^{uv}-\int\overbrace{\frac{2}{3}x(a^{2}+x^{2})^{\frac{3}{2}}dx}^{vdu}.$ We can integrate the last one by substitution. Let $$t=a^{2}+x^{2}.$$ Then $$dt=2xdx$$ and \begin{aligned} \frac{2}{3}\int x(a^{2}+x^{2})^{\frac{2}{3}}dx & =\frac{2}{3}\int t^{\frac{3}{2}}\cdot\overbrace{\frac{1}{2}dt}^{xdx}\\ & =\frac{1}{3}\cdot\frac{2}{5}t^{5/2}+K\\ & =\frac{2}{15}(a^{2}+x^{2})^{\frac{5}{2}}+K.\end{aligned} Finally $\int x^{3}\sqrt{a^{2}+x^{2}}dx=\frac{1}{3}x^{2}(a^{2}+x^{2})^{\frac{3}{2}}-\frac{2}{15}(a^{2}+x^{2})^{\frac{5}{2}}+C.$

Example 14

Evaluate $${\displaystyle \int\frac{x^{2}}{(a^{2}-x^{2})^{\frac{3}{2}}}dx}$$ ($$a>0$$).

Solution 14

We may try $u=x,\qquad dv=\frac{x}{(a^{2}-x^{2})^{\frac{3}{2}}},$ because we can integrate $$dv$$ by making the substitution $$z=a^{2}-x^{2}$$ and $$dx=-2xdx$$: \begin{aligned} \int\frac{x}{(a^{2}-x^{2})^{\frac{3}{2}}}dx & =\int\frac{-dz}{2z^{\frac{3}{2}}}=-\frac{1}{2}\int z^{-\frac{3}{2}}dz\\ & =-\frac{1}{2}(-2)z^{-\frac{1}{2}}+C_{1}\\ & =\frac{1}{\sqrt{z}}+C_{1}\\ & =\frac{1}{\sqrt{a^{2}-x^{2}}}+C_{1}.\end{aligned} So $u=x\Rightarrow du=dx$ $dv=\frac{x}{(a^{2}-x^{2})^{\frac{3}{2}}}\Rightarrow v=\frac{1}{\sqrt{a^{2}-x^{2}}},$ and $\int\frac{x^{2}}{(a^{2}-x^{2})^{\frac{3}{2}}}dx=\overbrace{\frac{x}{\sqrt{a^{2}-x^{2}}}}^{uv}-\int\overbrace{\frac{1}{\sqrt{a^{2}-x^{2}}}}^{v}\overbrace{dx}^{du}.$ Because \begin{aligned} \int\frac{1}{\sqrt{a^{2}-x^{2}}}dx & =\frac{1}{a}\int\frac{1}{\sqrt{1-\left(\dfrac{x}{a}\right)^{2}}}dx\\ & =\frac{1}{a}\int\frac{\overbrace{adt}^{dx}}{\sqrt{1-t^{2}}}dt && \small{(t=\frac{x}{a},dt=\frac{1}{a}dx)}\\ & =\arcsin t+K\\ & =\arcsin\frac{x}{a}+K,\end{aligned} we get$\int\frac{x^{2}}{(a^{2}-x^{2})^{\frac{3}{2}}}dx=\frac{x}{\sqrt{a^{2}-x^{2}}}-\arcsin\frac{x}{a}+C,$ where $$C=-K$$.

### Tabular Integration by Parts

Consider an integral of the form $\int f(x)g(x)dx,$ where $$f(x)$$ can be differentiated repeatedly until its derivative becomes zero and $$g(x)$$ is a function that can be integrated repeatedly without difficulty. Such an integral can be evaluated by integration by parts. However, we may need to repeat integration by parts several times, which can be cumbersome. To organize the calculations and to expedite this process, we make a two-column table. In one column we write $$f(x)$$ and its derivatives, and in one column we write $$g(x)$$ and its integrals, then we multiply these and add them together in a fashion that is illustrated in the following examples.

Example 15

Evaluate $${\displaystyle \int x^{2}\sin x\ dx}$$.

Solution 15

Let $$f(x)=x^{2}$$ and $$g(x)=\sin x$$:

Thus $\int x^{2}\sin xdx=-x^{2}\cos x+2x\sin x+2\cos x+C.$

In the end, don’t forget to add $$C$$ as the constant of integration.

Example 16

Evaluate $${\displaystyle \int x^{3}e^{x}dx}$$.

Solution 16

Let $$f(x)=x^{3}$$ and $$g(x)=e^{x}.$$

Thus $\int x^{3}e^{x}dx=x^{3}e^{x}-3x^{2}e^{x}+6x\,e^{x}-6e^{x}+C.$

Example 17

Evaluate $${\displaystyle \int x^{3}\sqrt{1+x}}\,dx.$$

Solution 17

Let $$f(x)=x^{3}$$ and $$g(x)=\sqrt{1+x}$$ because the higher derivatives of $$f(x)$$ become zero and we can easily integrate $$g(x)$$ as many times as we wish.

Therefore, \begin{aligned} \int x^{3}\sqrt{1+x}\,dx & =\frac{2}{3}x^{3}(1+x)^{2/3}-\frac{4}{5}x^{2}(1+x)^{5/2}\\ & \qquad+\frac{16}{35}x(1+x)^{7/2}-\frac{32}{315}(1+x)^{9/2}+C.\end{aligned}

### Solving for Unknown Integral

Sometimes to evaluate an integral, we need to integrate by parts twice and then solve for the unknown integral.

Example 18

Evaluate $${\displaystyle \int e^{x}\sin x\,dx}.$$

Solution 18

Let $u=\sin x,\qquad dv=e^{x}dx;$ then $du=\cos x\,dx,\qquad v=e^{x}.$ Integration by parts produces $\int\overbrace{\sin x}^{u}\overbrace{e^{x}dx}^{dv}=\overbrace{\sin x\,e^{x}}^{uv}-\int\overbrace{e^{x}}^{v}\overbrace{\cos x\:dx}^{du}.\tag{i}$ The last integral is similar to the original integral, except it has $$\cos x$$ instead of $$\sin x$$. To find this integral, we need to integrate by parts with $U=\cos x,\qquad dV=e^{x}dx.$ So $dU=-\sin x\,dx,\qquad V=e^{x}$ and $\int\overbrace{\cos x}^{U}\overbrace{e^{x}dx}^{dV}=\overbrace{\cos x\:e^{x}}^{UV}-\int\overbrace{e^{x}}^{V}\overbrace{(-\sin x)dx}^{dU}.\tag{ii}$ Substituting (ii) in (i) yields $\int\sin x\,e^{x}dx=e^{x}\sin x-e^{x}\cos x-\int\sin x\,e^{x}dx$ Here the original integral appears on the left and the right hand sides. $2\int e^{x}\sin xdx=-e^{x}\cos x+e^{x}\sin x+C_{1}.$ Here we add the constant of integration to cover all the possible integral functions (or antiderivatives). Finally $\int e^{x}\sin xdx=\frac{1}{2}e^{x}\left(\sin x-\cos x\right)+C_{2},$ with $$C_{2}=C_{1}/2$$.

Notice that we could start with $$u=e^{x}$$ and $$dv=\sin x\:dx$$. $u=e^{x},\qquad\qquad dv=\sin x\,dx;$ then $du=e^{x}dx,\qquad\qquad v=-\cos x$ so that $\int e^{x}\sin xdx=-e^{x}\cos x-\int(-\cos x)e^{x}dx\tag{i}$ The last integral is like the original integral, except it has $$\cos x$$ instead of $$\sin x$$. To find this integral, we need to integrate by parts with $U=e^{x},\qquad\qquad dV=\cos x\,dx.$ Therefore, $dU=e^{x}dx,\qquad\qquad V=-\sin x$ and $\int\cos x\,e^{x}dx=e^{x}\sin x-\int\sin x\,e^{x}dx.\tag{ii}$ Substituting (ii) in (i), we get $\int e^{x}\sin xdx=-e^{x}\cos x+e^{x}\sin x-\int e^{x}\sin xdx$ Here the original integral appears on the left and the right hand sides. $2\int e^{x}\sin xdx=-e^{x}\cos x+e^{x}\sin x+C_{1}.$ Here we add the constant of integration to cover all the possible integral functions (or antiderivatives). Finally $\int e^{x}\sin xdx=-\frac{1}{2}e^{x}\cos x+\frac{1}{2}e^{x}\sin x+C_{2},$ with $$C_{2}=C_{1}/2$$.

• Sometimes, we need to combine integration by parts and integration by substitution, and then solve for the unknown function.

Example 19

Evaluate $${\displaystyle \int\sec^{3}x\,dx}.$$

Solution

This may be divided into two factors $u=\sec x,\qquad dv=\sec^{2}x\,dx.$ $$u$$ can be easily differentiated, and $$dv$$ can be easily integrated. So $du=\sec x\,\tan x\,dx,\qquad v=\tan x,$ and $\int\sec^{3}x\,dx=\overbrace{\sec x\ \tan x}^{uv}-\int\overbrace{\tan^{2}x\,\sec x\,dx}^{vdu}.$ Because $$1+\tan^{2}x=\sec^{2}x$$, we have \begin{aligned} \int\sec^{3}x\,dx & =\sec x\,\tan x-\int(\sec^{2}x-1)\sec x\,dx\\ & =\sec x\,\tan x-\int\sec^{3}x\,dx-\int\sec x\,dx,\end{aligned} or $2\int\sec^{3}x\,dx=\sec x\,\tan x-\int\sec x\,dx$ Previously we showed that $\int\sec x\,dx=\ln|\sec x+\tan x|+K.$ Therefore, $\int\sec^{3}x\,dx=\frac{1}{2}\sec x\:\tan x+\frac{1}{2}\ln|\sec x+\tan x|+C.$

1. Reference: Kasube, Herbert E. “A technique for integration by parts.” The American Mathematical Monthly 90.3 (1983): 210-211.↩︎