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 Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=1+x^{3}} . Then, $du=3x^{2}dx$ and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {du}{3}}=x^{2}dx} .
Therefore, the integral becomes  ${\frac {1}{3}}\int {\sqrt {u}}~du$.

Step 2:
We now have:
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 $u$-substitution. Let Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle u=\sin(x)} . Then, Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=\cos(x)dx} . Also, we need to change the bounds of integration.
Plugging in our values into the equation 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=\sin(x)} , we get 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=\sin\bigg(\frac{\pi}{4}\bigg)=\frac{\sqrt{2}}{2}} 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 u_2=\sin\bigg(\frac{\pi}{2}\bigg)=1} .
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 \int_{\frac{\sqrt{2}}{2}}^1 \frac{1}{u^2}~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 _{\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)   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{2}{9}(1+x^3)^{\frac{3}{2}}+C}
(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 -1+\sqrt{2}}

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