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, $\int {\frac {\ln x}{x}}~dx=\int u~du={\frac {u^{2}}{2}}+C={\frac {(\ln x)^{2}}{2}}+C$.

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:
$\int x^{2}{\sqrt {1+x^{3}}}~dx={\frac {1}{3}}\int {\sqrt {u}}~du={\frac {2}{9}}u^{\frac {3}{2}}+C={\frac {2}{9}}(1+x^{3})^{\frac {3}{2}}+C$.

(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:
$\int _{\frac {\pi }{4}}^{\frac {\pi }{2}}{\frac {\cos(x)}{\sin ^{2}(x)}}~dx=\int _{\frac {\sqrt {2}}{2}}^{1}{\frac {1}{u^{2}}}~du=\left.{\frac {-1}{u}}\right|_{\frac {\sqrt {2}}{2}}^{1}=-{\frac {1}{1}}-{\frac {-1}{\frac {\sqrt {2}}{2}}}=-1+{\sqrt {2}}$.

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

Return to Sample Exam