Q106 / 160 · AP EAMCET 2015 · Mathematics

$\int_0^{\frac{\pi}{4}} \frac{\sin x+\cos x}{3+\sin 2 x} d x$ is equal to

  • A
    $\frac{1}{2} \log 3$
  • B
    $\log 2$
  • C
    $\log 3$
  • Correct answer
    D
    $\frac {1}{4} \log 3$

Solution

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, $I=\int_{-1}^0 \frac{d t}{3+1-t^2}=\int_{-1}^0 \frac{d t}{(2)^2-(t)^2}$ $\begin{aligned} & =\frac{1}{4}\left[\log \left(\frac{2+t}{2-t}\right)\right]_{-1}^0 \\ & =-\frac{1}{4} \log \frac{1}{3}=\frac{1}{4} \log 3 \end{aligned}$

Hence, the correct answer is option: 'D'.

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