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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS4 Marks
Question
Evaluate: $\int\limits^{\frac{\pi}{2}}_{0}\frac{\sin^{2}x}{\sin x + \cos x}dx$
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Answer

$\text{Let I} = \int\limits^{\pi/2}_{0}\frac{\sin^{2}\text{x}}{\sin\text{x} + \cos \text{x}}\text{dx Also I} = \int\limits^{\pi/2}_{0}\frac{\sin^{2}\bigg(\frac{\pi}{2} - \text{x}\bigg)}{\sin\bigg(\frac{\pi}{2} - \text{x}\bigg) + \cos \bigg(\frac{\pi}{2} - \text{x}\bigg)} \text{dx} = \int\limits^{\pi/2}_{0}\frac{\cos^{2}\text{x}}{\cos \text{x} + \sin \text{x}}\text{dx}$
Adding to get, $\text{2I} = \int\limits^{\pi/2}_{0}\frac{1}{\sin\text{x} + \cos\text{x}}\text{dx}= \frac{1}{\sqrt{2}}\int\limits^{\pi/2}_{0}\frac{1}{\cos(\text{x} -\pi/4}\text{dx}$
$\Rightarrow\text{2I} = \frac{1}{\sqrt{2}}\int\limits^{\pi/2}_{0}\sec(\text{x} - \pi/4)\text{dx} = \frac{1}{\sqrt{2}}\log\Bigg|\sec\bigg(\text{x} - \frac{\pi}{4}\bigg) + \tan\bigg(\text{x} - \frac{\pi}{4}\bigg)\Bigg|^{\pi/2}_{0}$
$\Rightarrow \text{2I} = \frac{1}{\sqrt{2}}\left\{\log|\sqrt{2} + 1 | -\log|\sqrt{2} - 1|\right\}$
$\Rightarrow\text{I} = \frac{1}{2\sqrt{2}}\bigg\{\log|\sqrt{2} + 1 | -\log|\sqrt{2} - 1| \ \ \ \text{or}\ \ \ \frac{1}{2\sqrt{2}}\log \bigg|\frac{\sqrt{2}+1}{\sqrt{2} - 1}\bigg|$

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