Q76 / 170 · Manipal (MET) 2016 · Mathematics

The value of $\int \frac{\sin x+\cos x}{3+\sin 2 x} d x$ is

  • Correct answer
    A
    $\frac{1}{4} \log \left(\frac{2-\sin x+\cos x}{2+\sin x-\cos x}\right)+C$
  • B
    $\frac{1}{2} \log \left(\frac{2+\sin x}{2-\sin x}\right)+C$
  • C
    $\frac{1}{4} \log \left(\frac{1+\sin x}{1-\sin x}\right)+C$
  • D
    None of the above

Solution

Let $I=\int \frac{\sin x+\cos x}{3+\sin 2 x} d x$ $=\int \frac{\sin x+\cos x}{4+\sin 2 x-1} d x$ $\Rightarrow I=\int \frac{\sin x+\cos x}{4-(\sin x-\cos x)^2} d x$ Put $\sin x-\cos x=t$ $\begin{aligned} & \Rightarrow \quad(\cos x+\sin x) d x=d t \\ & \therefore \quad I=\int \frac{d t}{4-t^2}=\frac{1}{4} \log \frac{2-t}{2+t}+C\end{aligned}$ $=\frac{1}{4} \log \left(\frac{2-\sin x+\cos x}{2+\sin x-\cos x}\right)+C$

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

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