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:
u substitution

Solution:

(a)

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

Step 2:
We integrate to get
$\int x^{2}\sin(x^{3})dx={\frac {-1}{3}}\cos(u)+C={\frac {-1}{3}}\cos(x^{3})+C$

(b)

Step 1:
Again, 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:
So, we get
$\int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\cos ^{2}(x)\sin(x)dx=\int _{\frac {\sqrt {2}}{2}}^{\frac {\sqrt {2}}{2}}-u^{2}=\left.{\frac {-u^{3}}{3}}\right|_{\frac {\sqrt {2}}{2}}^{\frac {\sqrt {2}}{2}}=0$

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

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