EXERCISE 7.2

Integration by parts technique is used to solve problems in a simple way with complete steps.

$1-38$ Evaluate the integral.

  1. $$\int x e^{-2 x} d x$$

Integrating by parts

$$
\int u v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x
$$

Here $u=x, v=e^{-2 x}$

$$
\begin{aligned}
\therefore \quad \int x e^{-2 x} d x & =x \int e^{-2 x} d x-\int\left[\frac{d(x)}{d x} \int e^{-2 x} d x\right] d x \\ \\
& =x \frac{e^{-2 x}}{-2}-\int(1) \frac{e^{-2 x}}{-2} d x \\ \\
& =-\frac{x e^{-2x}}{2}+\frac{1}{2} \int e^{-2 x} d x \\ \\
& =-\frac{x e^{-2 x}}{2}+\frac{1}{2}\left[\frac{e^{-2 x}}{-2}\right]+c \\ \\
& =-\frac{x e^{-2 x}}{2}-\frac{1}{4} e^{-2 x}+c
\end{aligned}
$$

2. $$\int x e^{3 x} d x$$

Here $$u=x,  \quad v=e^{3 x}$$

Formula integration by parts:

$$\int u v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x$$

integrating by parts

$$
\begin{aligned}

& \int x e^{3 x} d x=x \int e^{3x} d x-\int\left[\frac{d}{d x}(x) \int e^{3 x} d x\right] d x \\ \\
&=x \frac{e^{3x}}{3}-\int(1) \frac{e^{3x}}{3} d x \\ \\ &=\frac{x e^{3x}}{3}-\frac{1}{3} \int e^{3 x} d x \\ \\
&=\frac{x e^{3x}}{3}-\frac{1}{3}\left[\frac{e^{3x}}{3}\right]+c \\ \\ &=\frac{x e^{3x}}{3}-\frac{1}{9} e^{3x}+c
\end{aligned}

$$

3.

$$
\begin{aligned}
& \int x^{2} e^{x} d x \\ \\
& u=x^{2}, \quad v=e^{x} \\ \\

& \text {Formula integration by parts:} \quad \int u v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x
\end{aligned}
$$

Integrating by parts

$$
\begin{aligned}
\int x^{2} e^{x} d x & =x^{2} \int e^{x} d x-\int\left[\frac{d\left(x^{2}\right)}{d x} \int e^{x} d x\right] d x \\ \\
& =x^{2} e^{x}-\int 2 x e^{x} d x
\end{aligned}
$$

Again integrating by parts

$$
\begin{aligned}
& =x^{2} e^{x}-2\left[x \int e^{x} d x-\int\left(\frac{d}{d x}(x) \int e^{x} d x\right) d x\right] \\ \\
= & x^{2} e^{x}-2 x e^{x}+2 \int(1) e^{x} d x \\ \\
= & x^{2} e^{x}-2 x e^{x}+2 e^{x}+c
\end{aligned}
$$

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

$$
u=x^{2}, \quad v=e^{-2 x}$$

Formula integration by parts:

$$\int u v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x$$

Integrating by parts

$$
\begin{aligned}
& \int x^{2} e^{-2 x} d x=x^{2} \int e^{-2 x} d x-\int\left[\frac{d}{d x}\left(x^{2}\right) \int e^{-2 x} d x\right] dx \\ \\
& =x^{2} \frac{e^{-2 x}}{-2}-\int 2 x \frac{e^{-2 x}}{-2} d x \\ \\
& =-\frac{x^{2} e^{-2 x}}{2}+\int x e^{-2 x} d x
\end{aligned}
$$

Again integrating by parts

$$

\begin{aligned}

& =-\frac{x^{2} e^{-2 x}}{2}+x \int e^{-2 x} d x-\int \left[\frac{d}{d x}\left(x\right) \int e^{-2 x} d x\right]dx \\ \\

&=-\frac{x^{2} e^{-2 x}}{2}+x \frac{e^{-2x}}{-2}- \int (1)\frac{e^{-2x}}{-2} dx \\ \\

&=-\frac{x^{2} e^{-2 x}}{2}- \frac{xe^{-2x}}{2} +\frac{1}{2} \int e^{-2x} dx \\ \\

&=-\frac{x^{2} e^{-2 x}}{2}- \frac{xe^{-2x}}{2} +\frac{1}{2} \left(\frac{e^{-2x}}{-2}\right)+c \\ \\

&=-\frac{x^{2} e^{-2 x}}{2}- \frac{xe^{-2x}}{2}-\frac{1}{4} e^{-2x}+c

\end{aligned}

$$

5.

$$
\begin{aligned}
& \int x \sin 3x d x \\ \\
& u=x, \quad v=\sin 3x \\ \\

& \text {Formula integration by parts:} \quad \int u v d x=u \int v d x-\int\left(\frac{d u}{d x} \int v d x\right) d x
\end{aligned}
$$

Integrating by parts

$$
\begin{aligned}

\int x \sin 3x d x= x \int \sin 3x dx-\int \left [ \frac {d}{dx}(x) \int\sin 3x dx \right]dx \\ \\

=x \left (\frac{-\cos 3x}{3}\right)-\int (1) \left(\frac{-\cos 3x}{3}\right)dx \\ \\

=-\frac{1}{3} x \cos 3x +\frac{1}{3} \int \cos 3x dx \\ \\

=-\frac{1}{3} x \cos 3x +\frac{1}{3} \left(\frac{\sin 3x}{3}\right)+c \\ \\

=-\frac{1}{3} x \cos 3x +\frac{1}{9} \sin 3x+c

\end{aligned}

$$