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

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

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