009B Sample Midterm 1, Problem 1

Evaluate the indefinite and definite integrals.

a) $\int x^{2}{\sqrt {1+x^{3}}}~dx$

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

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

Solution:

(a)

Step 1:
We need to use $u$-substitution. Let $u=1+x^{3}.$ Then, $du=3x^{2}dx$ and  ${\frac {du}{3}}=x^{2}dx.$
Therefore, the integral becomes
${\frac {1}{3}}\int {\sqrt {u}}~du.$

Step 2:
We now have:
${\begin{array}{rcl}\displaystyle {\int x^{2}{\sqrt {1+x^{3}}}~dx}&=&\displaystyle {\int x^{2}{\sqrt {1+x^{3}}}~dx}\\&&\\&=&\displaystyle {{\frac {1}{3}}\int {\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {2}{9}}u^{\frac {3}{2}}+C}\\&&\\&=&\displaystyle {{\frac {2}{9}}(1+x^{3})^{\frac {3}{2}}+C.}\\\end{array}}$

(b)

Step 1:
Again, we need to use $u$-substitution. Let $u=\sin(x).$ Then, $du=\cos(x)dx.$ Also, we need to change the bounds of integration.
Plugging in our values into the equation $u=\sin(x),$ we get $u_{1}=\sin {\bigg (}{\frac {\pi }{4}}{\bigg )}={\frac {\sqrt {2}}{2}}$ and $u_{2}=\sin {\bigg (}{\frac {\pi }{2}}{\bigg )}=1.$
Therefore, the integral becomes
$\int _{\frac {\sqrt {2}}{2}}^{1}{\frac {1}{u^{2}}}~du.$

Step 2:
We now have:
${\begin{array}{rcl}\displaystyle {\int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {\cos(x)}{\sin ^{2}(x)}}~dx}&=&\displaystyle {\int _{\frac {\sqrt {2}}{2}}^{1}{\frac {1}{u^{2}}}~du}\\&&\\&=&\displaystyle {\left.{\frac {-1}{u}}\right|_{\frac {\sqrt {2}}{2}}^{1}}\\&&\\&=&\displaystyle {-{\frac {1}{1}}-{\frac {-1}{\frac {\sqrt {2}}{2}}}}\\&&\\&=&\displaystyle {-1+{\sqrt {2}}.}\\\end{array}}$

Final Answer:
(a)   ${\frac {2}{9}}(1+x^{3})^{\frac {3}{2}}+C$
(b)   $-1+{\sqrt {2}}$

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