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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=x^{2}+1} . Then, $du=2xdx$.
Thus, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int 2x(x^{2}+1)^{3}~dx=\int u^{3}~du={\frac {u^{4}}{4}}+C={\frac {(x^{2}+1)^{4}}{4}}+C} .

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
$\int x^{2}\sin(x^{3})~dx={\frac {-1}{3}}\cos(u)+C={\frac {-1}{3}}\cos(x^{3})+C$

(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
$\int _{-{\frac {\pi }{4}}}^{\frac {\pi }{4}}\cos ^{2}(x)\sin(x)~dx=\int _{\frac {\sqrt {2}}{2}}^{\frac {\sqrt {2}}{2}}-u^{2}~du=\left.{\frac {-u^{3}}{3}}\right|_{\frac {\sqrt {2}}{2}}^{\frac {\sqrt {2}}{2}}=0$

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

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