Evaluate the following integrals: ^ - 2 _ -3 2 ^2(3 +x)+( +x)^3 dx

Let I= ^ - 2 _ -3 2 ^2(3 +x)+( +x)^3 dx Put +x=z =dz When x -3 2 , z - 2 When x - 2 , z 2 I= ^ - 2 _ - 2 [ ^2(2 +z)+z^3 ]dx = ^ - 2 _ - 2 ( ^2z+z^3 )dz [ (2 + )= ] =1 2 ^ - 2 _ - 2 1- 2z 2 dz+ ^ - 2 _ - 2 z^3 dz =1 2 ^ - 2 _ - 2 dz-1 2 ^ - 2 _ - 2 2z dz+ ^ - 2 _ - 2 z^3 dz =1 2 [z ]^ 2 _ - 2 -1 2 [ 2z 2 ]^ 2 _ - 2 + [ z^4 4 ]^ 2 _ - 2 =1 2 [ 2 - (- 2 ) ]-1 4 [ - (- ) ]+1 4 ( ^4 16 - ^4 16 ) =1 2 -1 4 (0+0)+1 4 0 = 2

Evaluate the following integrals: ^ - 2 _ -3 2 ^2(3 +x)+( +x)^3 d…

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Question

Evaluate the following integrals:$\int\limits^{-\frac{\pi}{2}}_{-\frac{3\pi}{2}}\Big\{\sin^2(3\pi+\text{x})+(\pi+\text{x})^3\Big\}\text{dx}$

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Answer

Let $\text{I}=\int\limits^{-\frac{\pi}{2}}_{-\frac{3\pi}{2}}\Big\{\sin^2(3\pi+\text{x})+(\pi+\text{x})^3\Big\}\text{dx}$ Put $\pi+\text{x}=\text{z}$$\Rightarrow\text{dx}=\text{dz}$
When $\text{x}\rightarrow-\frac{3\pi}{2},\text{ z}\rightarrow-\frac{\pi}{2}$ When $\text{x}\rightarrow-\frac{\pi}{2},\text{ z}\rightarrow\frac{\pi}{2}$$\therefore\ \text{I}=\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\Big[\sin^2(2\pi+\text{z})+\text{z}^3\Big]\text{dx}$
$=\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\big(\sin^2\text{z}+\text{z}^3\big)\text{dz}$ $\big[\sin(2\pi+\theta)=\sin\theta\big]$
$=\frac{1}{2}\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\frac{1-\cos2\text{z}}{2}\text{ dz}+\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{z}^3\text{ dz}$
$=\frac{1}{2}\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{dz}-\frac{1}{2}\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\cos2\text{z dz}+\int\limits^{-\frac{\pi}{2}}_{-\frac{\pi}{2}}\text{z}^3\text{ dz}$
$=\frac{1}{2}\Big[\text{z}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}-\frac{1}{2}\Big[\frac{\sin2\text{z}}{2}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}+\Big[\frac{\text{z}^4}{4}\Big]^{\frac{\pi}{2}}_{-\frac{\pi}{2}}$
$=\frac{1}{2}\bigg[\frac{\pi}{2}-\Big(-\frac{\pi}{2}\Big)\bigg]-\frac{1}{4}\big[\sin\pi-\sin(-\pi)\big]+\frac{1}{4}\Big(\frac{\pi^4}{16}-\frac{\pi^4}{16}\big)$
$=\frac{1}{2}\times\pi-\frac{1}{4}(0+0)+\frac{1}{4}\times0$
$=\frac{\pi}{2}$

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