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INTEGRALS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsINTEGRALS3 Marks
Question
Evaluate the definite integral in Exercise:
$\int\limits^{\frac{\pi}{4}}_0\frac{\sin\text{x}\cos\text{x}}{\cos^{4}\text{x}+\sin^{4}\text{x}}\text{dx}$
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Answer

$\text{Let I}=\int^{\frac{\pi}{4}}\limits_{0}\frac{\sin\text{x}\cos\text{x}}{\cos^{4}\text{x}+\sin^{4}\text{x}}\text{dx}$
$\Rightarrow\text{ I}=\int^{\frac{\pi}{4}}\limits_{0}\frac{\frac{(\sin\text{x}\cos\text{x)}}{\cos^{4}\text{x}}}{\frac{\cos^{4}\text{x}+\sin^{4}\text{x}}{\cos^{4}\text{x}}}\text{dx}$
$\Rightarrow\text{ I}=\int^{\frac{\pi}{4}}\limits_{0}\frac{\tan\text{x}\sec^{2}\text{x}}{1+\tan^{4}\text{x}}\text{dx}$
$\text{Let}\ \tan^{2}\text{x}=\text{t}\ \Rightarrow2\tan\text{x}\text{dx}=\text{dt}$
when $\text{x}=0,\text{t}=0$ and when $\text{x}=\frac{\pi}{4},\text{t}=1$
$\therefore\ \text{I}=\frac{1}{2}\int^{1}\limits_{0}\frac{\text{dt}}{1+\text{t}^{2}}$
$=\frac{1}{2}\Big[\tan^{-1}\text{t}\Big]^{1}_{0}$
$=\frac{1}{2}\Big[\tan^{-1}-\tan^{-1}0\Big]$
$=\frac{1}{2}\Big[\frac{\pi}{4}\Big]$
$=\frac{\pi}{8}$

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