009B Sample Midterm 3, Problem 3

Compute the following integrals:

(a)   $\int x^{2}\sin(x^{3})~dx$

(b)   $\int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\cos ^{2}(x)\sin(x)~dx$

Foundations:
How would you integrate $2x(x^{2}+1)^{3}~dx?$
You could use $u$-substitution.
Let $u=x^{2}+1.$
Then, $du=2x~dx.$
Thus,
${\begin{array}{rcl}\displaystyle {\int 2x(x^{2}+1)^{3}~dx}&=&\displaystyle {\int u^{3}~du}\\&&\\&=&\displaystyle {{\frac {u^{4}}{4}}+C}\\&&\\&=&\displaystyle {{\frac {(x^{2}+1)^{4}}{4}}+C.}\\\end{array}}$

Solution:

(a)

Step 1:
We proceed using $u$-substitution.
Let $u=x^{3}.$
Then, $du=3x^{2}~dx$ and ${\frac {du}{3}}=x^{2}~dx.$
Therefore, we have
$\int x^{2}\sin(x^{3})~dx=\int {\frac {\sin(u)}{3}}~du.$

Step 2:
We integrate to get
${\begin{array}{rcl}\displaystyle {\int x^{2}\sin(x^{3})~dx}&=&\displaystyle {{\frac {-1}{3}}\cos(u)+C}\\&&\\&=&\displaystyle {{\frac {-1}{3}}\cos(x^{3})+C.}\\\end{array}}$

(b)

Step 1:
We proceed using u substitution.
Let $u=\cos(x).$
Then, $du=-\sin(x)~dx.$
Since this is a definite integral, we need to change the bounds of integration.
We have $u_{1}=\cos {\bigg (}-{\frac {\pi }{4}}{\bigg )}={\frac {\sqrt {2}}{2}}$ and $u_{2}=\cos {\bigg (}{\frac {\pi }{4}}{\bigg )}={\frac {\sqrt {2}}{2}}.$

Step 2:
Therefore, we get
${\begin{array}{rcl}\displaystyle {\int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\cos ^{2}(x)\sin(x)~dx}&=&\displaystyle {\int _{\frac {\sqrt {2}}{2}}^{\frac {\sqrt {2}}{2}}-u^{2}~du}\\&&\\&=&\displaystyle {\left.{\frac {-u^{3}}{3}}\right|_{\frac {\sqrt {2}}{2}}^{\frac {\sqrt {2}}{2}}}\\&&\\&=&\displaystyle {0.}\\\end{array}}$

Final Answer:
(a)     ${\frac {-1}{3}}\cos(x^{3})+C$
(b)     $0$

Return to Sample Exam