Evaluate the indefinite and definite integrals.
- a)
- 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_0^\pi \sin^2x~dx}
| Foundations:
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| Review 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}
-substitution
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| Trig identities
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Solution:
(a)
| Step 1:
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| We start by writing 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 \tan^3x~dx=\int \tan^2x\tan x ~dx}
.
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| Since 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 \tan^2x=\sec^2x-1}
, we 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 \tan^3x~dx=\int (\sec^2x-1)\tan x ~dx=\int \sec^2\tan x~dx-\int \tan x~dx}
.
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| Step 2:
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| Now, we need to use 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}
-substitution for the first integral. 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=\tan(x)}
. 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=\sec^2xdx}
. So, we have
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| 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 \tan^3x~dx=\int u~du-\int \tan x~dx=\frac{u^2}{2}-\int \tan x~dx=\frac{\tan^2x}{2}-\int \tan x~dx}
.
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| Step 3:
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| For the remaining integral, we also need to use 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}
-substitution. First, we write 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 \tan^3x~dx=\frac{\tan^2x}{2}-\int \frac{\sin x}{\cos x}~dx}
.
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| Now, we 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=\cos x}
. 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=-\sin xdx}
. So, we get
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| 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 \tan^3x~dx=\frac{\tan^2x}{2}+\int \frac{1}{u}~dx=\frac{\tan^2x}{2}+\ln |u|+C=\frac{\tan^2x}{2}+\ln |\cos x|+C}
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(b)
| Step 1:
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One of the double angle formulas is 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 \cos(2x)=1-2\sin^2(x)}
. Solving for  , we get  .
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| Plugging this identity into our integral, we get Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{\pi }\sin ^{2}x~dx=\int _{0}^{\pi }{\frac {1-\cos(2x)}{2}}~dx=\int _{0}^{\pi }{\frac {1}{2}}~dx-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx}
.
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| Step 2:
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| If we integrate the first integral, we get
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| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{\pi }\sin ^{2}x~dx=\left.{\frac {x}{2}}\right|_{0}^{\pi }-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx={\frac {\pi }{2}}-\int _{0}^{\pi }{\frac {\cos(2x)}{2}}~dx}
.
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| Step 3:
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For the remaining integral, we need to use  -substitution. Let  . Then,  and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {du}{2}}=dx}
. Also, since this is a definite integral
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and we are using -substitution, we need to change the bounds of integration. We have  and  .
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| So, the integral becomes
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| Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int _{0}^{\pi }\sin ^{2}x~dx={\frac {\pi }{2}}-\int _{0}^{2\pi }{\frac {\cos(u)}{4}}~du={\frac {\pi }{2}}-\left.{\frac {\sin(u)}{4}}\right|_{0}^{2\pi }={\frac {\pi }{2}}-{\bigg (}{\frac {\sin(2\pi )}{4}}-{\frac {\sin(0)}{4}}{\bigg )}={\frac {\pi }{2}}}
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| Final Answer:
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(a) 
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(b) 
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