009B Sample Midterm 1, Problem 1

Evaluate the indefinite and definite integrals.

a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^2\sqrt{1+x^3}dx}

b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int _{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\cos(x)}{\sin^2(x)}dx}

Foundations:

Solution:

(a)

Step 1:
We need to use substitution. Let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle u=1+x^3} . Then, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle du=3x^2dx} and Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{du}{3}=x^2dx} .
Therefore, the integral becomes Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{3}\int \sqrt{u}du} .

Step 2:
We now have:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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) $-1+{\sqrt {2}}$

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