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Question: Answered & Verified by Expert
$\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{3+\sin 2 x} d x$ is equal to
MathematicsDefinite IntegrationAP EAMCETAP EAMCET 2015
Options:
  • A $\frac{1}{2} \log 3$
  • B $\log 2$
  • C $\log 3$
  • D $\frac {1}{4} \log 3$
Solution:
1690 Upvotes Verified Answer
The correct answer is: $\frac {1}{4} \log 3$
Let $I=\int_0^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$
Put $\sin x-\cos x=t$
$(\sin x+\cos x) d x=d t$
Limits:
$\begin{aligned}
& x=0 \\
& \Rightarrow t=\sin \theta-\cos \theta=-1 \\
& x=\frac{\pi}{4} \Rightarrow t=\sin \frac{\pi}{4}-\cos \frac{\pi}{4}=0 \\
& \sin 2 x=1-1+\sin 2 x \\
& =1-(1-\sin 2 x) \\
& =1-\left(\sin ^2 x+\cos ^2 x-2 \sin x \cos x\right) \\
& =1-(\sin x-\cos x)^2 \\
& =1-t^2
\end{aligned}$
Now, $I=\int_{-1}^0 \frac{d t}{3+1-t^2}=\int_{-1}^0 \frac{d t}{(2)^2-(t)^2}$
$\begin{aligned}
& =\frac{1}{4}\left[\log \left(\frac{2+t}{2-t}\right)\right]_{-1}^0 \\
& =-\frac{1}{4} \log \frac{1}{3}=\frac{1}{4} \log 3
\end{aligned}$

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