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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate the following integrals:
$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}|\sin\text{x}|\text{dx}$
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Answer

$\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}}\sin\text{x}|\sin\text{x}|\text{dx}$
$=\int\limits^{0}_{-\frac{\pi}{4}}\sin\text{x}|\sin\text{x}|\text{dx}+\int\limits_{0}^{\frac{\pi}{2}}\sin\text{x}|\sin\text{x}|\text{dx}$
$=\int\limits^{0}_{-\frac{\pi}{4}}\sin\text{x}(-\sin\text{x})\text{dx}+\int\limits_{0}^{\frac{\pi}{2}}\sin\text{x}(\sin\text{x})\text{dx}$ $\begin{pmatrix}|\sin\text{x}|=\begin{cases}\sin\text{x},&0\leq\text{x}\leq\frac{\pi}{2}\\-\sin\text{x},&-\frac{\pi}{4}\leq\text{x}\leq0\end{cases} \end{pmatrix} $
$=\int\limits^{0}_{-\frac{\pi}{4}}\sin^2\text{x dx}+\int\limits_{0}^{\frac{\pi}{2}}\sin^2\text{x dx}$
$=-\int\limits^{0}_{-\frac{\pi}{4}}\frac{1-\cos2\text{x}}{2}\text{ dx}+\int\limits_{0}^{\frac{\pi}{2}}\frac{1-\cos2\text{x}}{2}\text{ dx}$
$=-\frac{1}{2}\int\limits^{0}_{-\frac{\pi}{4}}\text{dx}+\frac{1}{2}\int\limits^{0}_{-\frac{\pi}{4}}\cos2\text{x dx}+\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\text{dx}-\frac{1}{2}\int\limits_{0}^{\frac{\pi}{2}}\cos2\text{x dx}$
$=-\frac{1}{2}\times\big[\text{x}\big]^0_\frac{\pi}{4}+\frac{1}{2}\times\Big[\frac{\sin2\text{x}}{2}\Big]^0_{\frac{\pi}{2}}+\frac{1}{2}\times\big[\text{x}\big]^{\frac{\pi}{2}}_0-\frac{1}{2}\times\Big[\frac{\sin2\text{x}}{2}\Big]^{\frac{\pi}{2}}_0$
$=-\frac{1}{2}\Big(0+\frac{\pi}{4}\Big)+\frac{1}{4}(0+1)+\frac{\pi}{4}-\frac{1}{4}(0-0)$
$=\frac{\pi}{8}+\frac{1}{4}$

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