Integration by parts method.

If $u(x)$ and $v(x)$ are differentiable functions, then the following integration by parts formula holds: $$\int u(x)v^{\prime }(x{)}^{\prime },dx=u(x)v(x)-\int v(x)u^{\prime }(x)dx,$$

Alternatively, in a concise notation: $$\int udv=uv-\int vdu.$$

This formula is used in cases where the integrand $f(x)dx$ can be expressed as $u,dv$ such that $v=\int ,dv$ can be found and the resulting integral on the right-hand side $\int v,du$ is simpler than the original $\int u,dv$.

Some types of functions that are integrated using the integration by parts method and recommended partitions:

Type I:

$$\int P_n(x)\cos mxdx;$$

$$\int P_n(x)\sin nxdx;$$

$$\int P_n(x)a^{\alpha x}dx,$$

where $P_n(x)$ is a polynomial of degree $n$ in $x:$ $u=P_n(x),$ and $dv$ is everything else.

Type II:

$$\int P_n(x){\ln }^{m}xdx;$$

$$\int P_n(x)\arccos xdx;$$

$$\int P_n(x)\arcsin xdx;$$

$$\int P_n(x)arctgxdx;$$

$$\int P_n(x)arcctgxdx,$$

where $P_n(x)$ is a polynomial of degree $n$ in $x:$ $dv=P_n(x)dx,$ and $u$ is everything else.

Type III:

$$\int a^{\alpha x}\cos mxdx;$$

$$\int a^{\alpha x}\sin nxdx:$$

the partition is arbitrary.

It should be noted that for computing the integral, the integration by parts formula can be applied repeatedly.

Examples:

1. $\int \arccos x,dx.$

Solution.

$$\int \arccos xdx=\left[\begin{array}{l}u=\arccos x\Rightarrow du=-\frac{1}{\sqrt{1-x^2}}dx\\ dv=dx\Rightarrow v=x\end{array}\right]=$$ $$=\arccos x\cdot x-\int x(-\frac{1}{\sqrt{1-x^2}})dx=x\arccos x+\int \frac{xdx}{\sqrt{1-x^2}}.$$

Let's compute the integral obtained on the right-hand side:

$$\int \frac{xdx}{\sqrt{1-x^2}}=\left[\begin{array}{l}1-x^2=t\\ dt=-2xdx\Rightarrow xdx=\frac{dt}{-2}\end{array}\right]=$$ $$=\int \frac{dt}{-2\sqrt{t}}=-\frac{1}{2}\int t^{-1/2}dt=\frac{1}{2}\frac{t^{1/2}}{1/2}+C=\sqrt{t}+C=\sqrt{1-x^2}+C.$$

Therefore,

$$\int \arccos xdx=x\arccos x+\sqrt{1-x^2}+C.$$

Answer: $x\arccos x+\sqrt{1-x^2}+C.$

2. $\int x\ln xdx.$

Solution.

$$\int x\ln xdx=\left[\begin{array}{l}u=\ln x\Rightarrow du=\frac{1}{x}dx\\ dv=xdx\Rightarrow v=\frac{x^2}{2}\end{array}\right]=\ln x\cdot \frac{x^2}{2}-\int \frac{x^2}{2}\frac{1}{x}dx=$$ $$=\frac{x^2}{2}\ln x-\frac{1}{2}\int xdx=\frac{x^2}{2}\ln x-\frac{x^2}{4}+C.$$

Answer: $\frac{x^2}{4}(2\ln x-1)+C.$

3. $\int x^2\sin xdx.$

Solution.

$$\int x^2\sin xdx=\left[\begin{array}{l}u=x^2\Rightarrow du=2xdx\\ dv=\sin xdx\Rightarrow v=-\cos x\end{array}\right]=$$ $$=-x^2\cos x+2\int \cos x\cdot xdxdx=\left[\begin{array}{l}u=x\Rightarrow du=dx\\ dv=\cos xdx\Rightarrow v=\sin x\end{array}\right]=$$ $$=-x^2\cos x+2(x\sin x-\int \sin xdx)=-x^2\cos x+2x\sin x+2\cos x+C.$$

Answer: $-x^2\cos x+2x\sin x+2\cos x+C.$

4.$\int x^3{e}^{-x^2}dx.$

Solution.

$$\int x^3{e}^{-x^2}dx=\left[\begin{array}{l}u=x^2\Rightarrow du=2xdx\\ dv=x{e}^{-x^2}dx\Rightarrow v=\int x{e}^{-x^2}dx\end{array}\right].$$

$$\int x{e}^{-x^2}dx=\left[\begin{array}{l}t=-x^2\\ dt=-2xdx\end{array}\right]=-\frac{1}{2}\int e^{t}dt=-\frac{1}{2}{e}^{-x^2}+c.$$ Therefore, $$\int x^3{e}^{-x^2}dx=\left[\begin{array}{l}u=x^2\Rightarrow du=2xdx\\ dv=x{e}^{-x^2}dx\Rightarrow v=-\frac{1}{2}{e}^{-x^2}dx\end{array}\right]=$$ $$=-\frac{1}{2}{x}^2{e}^{-x^2}-\int (-\frac{1}{2}{e}^{-x^2})2xdx=-\frac{1}{2}{x}^2{e}^{-x^2}+\int x{e}^{-x^2}dx=$$ $$=-\frac{1}{2}{x}^2{e}^{-x^2}-\frac{1}{2}{e}^{-x^2}+C.$$

Answer: $-\frac{1}{2}{e}^{-x^2}(x^2+1)+C.$

5. $\int \frac{x\sin x}{{\cos }^{3}x}dx.$

Solution.

$$\int \frac{x\sin x}{{\cos }^{3}x}dx=\left[\begin{array}{l}u=x\Rightarrow du=dx\\ dv=\frac{\sin x}{{\cos }^{3}x}dx\Rightarrow v=\int \frac{\sin x}{{\cos }^{3}x}dx\end{array}\right].$$

$$\int \frac{\sin x}{{\cos }^{3}x}dx=\left[\begin{array}{l}t=\cos x\\ dt=-\sin xdx\end{array}\right]=-\int \frac{1}{t^3}dt=-\int t^{-3}dt=$$ $$=-\frac{t^{-2}}{-2}=\frac{1}{2{\cos }^{2}x}+c.$$

Therefore,

$$\int \frac{x\sin x}{{\cos }^{3}x}dx=\left[\begin{array}{l}u=x\Rightarrow du=dx\\ dv=\frac{\sin x}{{\cos }^{3}x}dx\Rightarrow v=\frac{1}{2{\cos }^{2}x}\end{array}\right]=$$ $$=x\cdot \frac{1}{2{\cos }^{2}x}-\int \frac{1}{2{\cos }^{2}x}dx=\frac{x}{2{\cos }^{2}x}-\frac{1}{2}tgx+C.$$

Answer: $\frac{x}{2{\cos }^{2}x}-\frac{1}{2}tgx+C.$

6.$\int e^{ax}\cos bxdx.$

Solution.

$$\int e^{ax}\cos bxdx=\left[\begin{array}{l}u=\cos bx\Rightarrow du=-b\sin bxdx\\ dv=e^{ax}dx\Rightarrow v=\frac{1}{a}{e}^{ax}\end{array}\right]=$$ $$=\cos bx\cdot \frac{1}{a}{e}^{ax}-\int \frac{1}{a}{e}^{ax}\cdot (-b\sin bx)dx=$$ $$=\frac{\cos bx}{a}{e}^{ax}+\frac{b}{a}\int e^{ax}\sin bxdx=\left[\begin{array}{l}u=\sin bx\Rightarrow du=b\cos bxdx\\ dv=e^{ax}dx\Rightarrow v=\frac{1}{a}{e}^{ax}\end{array}\right]=$$ $$=\frac{\cos bx}{a}{e}^{ax}+\frac{b}{a}(\sin bx\cdot \frac{1}{a}{e}^{ax}-\int \frac{1}{a}{e}^{ax}b\cos bxdx)=$$ $$=\frac{\cos bx}{a}{e}^{ax}+\frac{b\sin bx}{a^2}{e}^{ax}-\frac{b^2}{a^2}\int e^{ax}\cos bxdx.$$

Let $\int e^{ax}\cos bx,dx=I.$ Then we'll rewrite the obtained equation as follows:

$$I=\frac{\cos bx}{a}{e}^{ax}+\frac{b\sin bx}{a^2}{e}^{ax}-\frac{b^2}{a^2}I.$$ Let's solve this equation for $I:$

$$I+\frac{b^2}{a^2}I=\frac{\cos bx}{a}{e}^{ax}+\frac{b\sin bx}{a^2}{e}^{ax}\Rightarrow$$

$$\Rightarrow I(1+\frac{b^2}{a^2})=\frac{\cos bx}{a}{e}^{ax}+\frac{b\sin bx}{a^2}{e}^{ax}\Rightarrow$$

$$\Rightarrow I=\frac{\frac{\cos bx}{a}{e}^{ax}+\frac{b\sin bx}{a^2}{e}^{ax}}{1+\frac{b^2}{a^2}}=\frac{\frac{e^{ax}}{a^2}(a\cos bx+b\sin bx)}{\frac{a^2+b^2}{a^2}}=$$ $$=\frac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}.$$

Answer: $\frac{e^{ax}(a\cos bx+b\sin bx)}{a^2+b^2}+C.$

7.$\int \cos (\ln x)dx.$

Solution.

$$\int \cos (\ln x)dx=\left[\begin{array}{l}u=\cos (\ln x)\Rightarrow du=-\frac{1}{x}\sin (\ln x)dx\\ dv=dx\Rightarrow v=x\end{array}\right]=$$ $$=\cos (\ln x)x-\int x\cdot (-\frac{1}{x}\sin (\ln x))dx=x\cos (\ln x)+\int \sin (\ln x)dx.$$

$$\int \sin (\ln x)dx=\left[\begin{array}{l}u=\sin (\ln x)\Rightarrow du=\frac{1}{x}\cos (\ln x)dx\\ dv=dx\Rightarrow v=x\end{array}\right]=$$ $$=\sin (\ln x)x-\int x\cdot (\frac{1}{x}\cos (\ln x))dx=x\sin (\ln x)-\int \cos (\ln x)dx.$$

Thus, $$\int \cos (\ln x)dx=x\cos (\ln x)+x\sin (\ln x)-\int \cos (\ln x)dx.$$

Let $\int \cos (\ln x),dx=I.$ Then we'll write and solve the equation:

$$I=x\cos (\ln x)+x\sin (\ln x)-I\Rightarrow$$ $$\Rightarrow 2I=x\cos (\ln x)+x\sin (\ln x)\Rightarrow$$ $$\Rightarrow I=\frac{1}{2}(x\cos (\ln x)+x\sin (\ln x).$$

Answer: $\frac{1}{2}(x\cos (\ln x)+x\sin (\ln x))+C.$

Homework.

1. $\int x\cos xdx.$

2. $\int \frac{\ln x}{\sqrt[3]{x}}dx.$

3. $\int x^2{e}^{-x}dx.$

4. $\int \frac{{\ln }^{2}x}{x^2}dx.$

5. $\int xarctgxdx.$

6. $\int \ln (x+\sqrt{1+x^2})dx.$

7. $\int x^3\ln xdx.$

8. $\int (x^2-2x+3)\cos xdx.$

Tags: Integration by parts method, calculus, integral, mathematical analysis