Evaluate : _0^ 4 ( x+ x 3+ 2 x ) d x - Vidyadip

Question

Evaluate : $\int_0^{\frac{\pi}{4}}\left(\frac{\sin x+\cos x}{3+\sin 2 x}\right) d x$

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Answer

Let
$I=\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{3+\sin 2 x} d x$
Here putting $\sin x-\cos x= t$
$(\cos x+\sin x) d x=d t$
So if $x=\frac{\pi}{4}$ then $t=0$ and if $x=0$ then $t=-1$
$(\sin x-\cos x)^2=1-2 \sin x \cos x=1-\sin 2 x$
$\therefore \sin 2 x=1-(\sin x-\cos x)^2$
$=1-t^2$
$I =\int_{-1}^0 \frac{d x}{3+\left(1-t^2\right)}=\int_{-1}^0 \frac{d t}{4-t^2}$
$I=\int_{-1}^0 \frac{d t}{(2)^2-t^2}=\frac{1}{2 \times 2}\left(\log \frac{2+t}{2-t}\right)_{-1}^0$
$\quad\left(\because \int \frac{1}{a^2-x^2} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+c\right)$
$I=\frac{1}{4}\left(\log 1-\log \frac{1}{3}\right)$
$I=\frac{1}{4}\left(0-\log \frac{1}{3}\right)=\frac{1}{4} \log 3$

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