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Model Paper 4 — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsModel Paper 43 Marks
Question
Evaluate the definite integral $\int_0^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^4 x+\sin ^4 x} d x$
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Answer

$I=\int_0^{\pi / 4} \frac{\sin x \cdot \cos x}{\cos ^4 x+\sin ^4 x} d x$
Dividing Nr. and Dr. by $\cos^4 x$
$=\int_0^{\pi / 4} \frac{\frac{\sin x \cdot \cos x}{\cos ^4 z}}{\frac{\cos ^4 x}{\cos ^4 x}+\frac{\sin ^4 x}{\cos ^4 x}} d x$
$=\int_0^{\pi / 4} \frac{\tan x \cdot \sec ^2 x}{1+\tan ^4 x} d x$
$=\int_0^{\pi / 4} \frac{\tan x \cdot \sec ^2 x}{1+\left(\tan ^2 x\right)^2} d x$
Put $\tan ^2 x=t$
$2 \tan x \cdot \sec ^2 x d x=d t$
When $x =0, t =0$ and when $x=\frac{\pi}{4}, t=1$
$\therefore I=\frac{1}{2} \int_0^1 \frac{d t}{1+t^2}$
$=\frac{1}{2}\left[\tan ^{-1} t\right]_0^1$
$=\frac{1}{2} \cdot \frac{\pi}{4}=\frac{\pi}{8}$

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