# 009B Sample Final 3, Problem 5

Find the following integrals.

(a)  $\int x\cos(x)~dx$ (b)  $\int \sin ^{3}(x)\cos ^{2}(x)~dx$ Foundations:
1. Integration by parts tells us that
$\int u~dv=uv-\int v~du.$ 2. Since  $\sin ^{2}x+\cos ^{2}x=1,$ we have
$\sin ^{2}x=1-\cos ^{2}x.$ Solution:

(a)

Step 1:
To calculate this integral, we use integration by parts.
Let  $u=x$ and  $dv=\cos xdx.$ Then,  $du=dx$ and  $v=\sin x.$ Therefore, we have
$\int x\cos(x)~dx=x\sin x-\int \sin x~dx.$ Step 2:
Then, we integrate to get
$\int x\cos(x)~dx=x\sin x+\cos x+C.$ (b)

Step 1:
First, we use the identity  $\sin ^{2}x=1-\cos ^{2}x$ to get
${\begin{array}{rcl}\displaystyle {\int \sin ^{3}(x)\cos ^{2}(x)~dx}&=&\displaystyle {\int \sin ^{2}x(\cos ^{2}x)\sin x~dx}\\&&\\&=&\displaystyle {\int (1-\cos ^{2}x)(\cos ^{2}x)\sin x~dx.}\end{array}}$ Step 2:
Now, we use  $u$ -substitution.
Let  $u=\cos(x).$ Then,  $du=-\sin(x)dx$ and  $-du=\sin(x)dx.$ Therefore, we have
${\begin{array}{rcl}\displaystyle {\int \sin ^{3}(x)\cos ^{2}(x)~dx}&=&\displaystyle {\int (-1)(1-u^{2})u^{2}~du}\\&&\\&=&\displaystyle {\int u^{4}-u^{2}~du}\\&&\\&=&\displaystyle {{\frac {u^{5}}{5}}-{\frac {u^{3}}{3}}+C}\\&&\\&=&\displaystyle {{\frac {\cos ^{5}x}{5}}-{\frac {\cos ^{3}x}{3}}+C.}\end{array}}$ (a)    $x\sin x+\cos x+C$ (b)    ${\frac {\cos ^{5}x}{5}}-{\frac {\cos ^{3}x}{3}}+C$ 