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Definite Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDefinite Integrals2 Marks
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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