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The value of the integral $\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$
is equal to
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is equal to
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Verified Answer
The correct answer is:
$\frac{1}{4} \log _{e} 3$
Let $I=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$
$=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+2 \sin x \cos x} d x$
$=\int_{0}^{\pi / 4}-\frac{\sin x+\cos x}{(\sin x-\cos x)^{2}-4} d x$
Put $\sin x-\cos x=t$
$\Rightarrow \quad(\cos x+\sin x) d x=d t$
when $x=0 \Rightarrow t=-1$
and $x=\frac{\pi}{4} \Rightarrow t=0$
$\therefore$ Eq. (i) becomes, $I=-\int_{-1}^{0} \frac{d t}{t^{2}-4}=-\left.\frac{1}{4}\left[\log \mid \frac{t-2}{t+2}\right]\right|_{-1} ^{0}$
$=-\frac{1}{4}(\log 1-\log 3)$
$=-\frac{1}{4}(0-\log 3)=\frac{1}{4} \log 3$
$=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+2 \sin x \cos x} d x$
$=\int_{0}^{\pi / 4}-\frac{\sin x+\cos x}{(\sin x-\cos x)^{2}-4} d x$
Put $\sin x-\cos x=t$
$\Rightarrow \quad(\cos x+\sin x) d x=d t$
when $x=0 \Rightarrow t=-1$
and $x=\frac{\pi}{4} \Rightarrow t=0$
$\therefore$ Eq. (i) becomes, $I=-\int_{-1}^{0} \frac{d t}{t^{2}-4}=-\left.\frac{1}{4}\left[\log \mid \frac{t-2}{t+2}\right]\right|_{-1} ^{0}$
$=-\frac{1}{4}(\log 1-\log 3)$
$=-\frac{1}{4}(0-\log 3)=\frac{1}{4} \log 3$
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