009B Sample Final 3, Problem 5 Detailed Solution

Find the following integrals.

(a)  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int x\cos(x)~dx}

(b)  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int \sin ^{3}(x)\cos ^{2}(x)~dx}

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Background Information:
1. Integration by parts tells us that
$\int u~dv=uv-\int v~du.$
2. Since  $\sin ^{2}x+\cos ^{2}x=1,$  we have
$\sin ^{2}x=1-\cos ^{2}x.$

Solution:

(a)

Step 1:
To calculate this integral, we use integration by parts.
Let  $u=x$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle dv=\cos xdx.}
Then,  $du=dx$  and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle v=\sin x.}
Therefore, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int x\cos(x)~dx=x\sin x-\int \sin x~dx.}

Step 2:
Then, we integrate to get
$\int x\cos(x)~dx=x\sin x+\cos x+C.$

(b)

Step 1:
First, we use the identity  $\sin ^{2}x=1-\cos ^{2}x$  to get
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int \sin ^{3}(x)\cos ^{2}(x)~dx}&=&\displaystyle {\int \sin ^{2}x(\cos ^{2}x)\sin x~dx}\\&&\\&=&\displaystyle {\int (1-\cos ^{2}x)(\cos ^{2}x)\sin x~dx.}\end{array}}}

Step 2:
Now, we use  $u$-substitution.
Let  $u=\cos(x).$
Then,  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle du=-\sin(x)dx}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle -du=\sin(x)dx.}
Therefore, we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\int \sin ^{3}(x)\cos ^{2}(x)~dx}&=&\displaystyle {\int (-1)(1-u^{2})u^{2}~du}\\&&\\&=&\displaystyle {\int u^{4}-u^{2}~du}\\&&\\&=&\displaystyle {{\frac {u^{5}}{5}}-{\frac {u^{3}}{3}}+C}\\&&\\&=&\displaystyle {{\frac {\cos ^{5}x}{5}}-{\frac {\cos ^{3}x}{3}}+C.}\end{array}}}

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 x\sin x +\cos x+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 \frac{\cos^5 x}{5}-\frac{\cos^3 x}{3}+C}

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