007B Sample Midterm 1, Problem 5 Detailed Solution

Find the area bounded by  $y=\sin(x)$  and  $y=\cos(x)$  from  $x=0$  to  $x={\frac {\pi }{4}}.$

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Background Information:
How would you integrate   $\int {\frac {\ln x}{x}}~dx?$
You can 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 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
${\begin{array}{rcl}\displaystyle {\int x^{2}{\sqrt {1+x^{3}}}~dx}&=&\displaystyle {{\frac {1}{3}}\int {\sqrt {u}}~du}\\&&\\&=&\displaystyle {{\frac {2}{9}}u^{\frac {3}{2}}+C}\\&&\\&=&\displaystyle {{\frac {2}{9}}(1+x^{3})^{\frac {3}{2}}+C.}\end{array}}$

(b)

Step 1:
We 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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