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Question: Answered & Verified by Expert
$\int \frac{\sin x d x}{3+4 \cos ^2 x}=$
MathematicsIndefinite IntegrationJEE Main
Options:
  • A $\log \left(3+4 \cos ^2 x\right)+c$
  • B $\frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{\cos x}{\sqrt{3}}\right)+c$
  • C $\frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c$
  • D $\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c$
Solution:
1234 Upvotes Verified Answer
The correct answer is: $\frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c$
$\begin{aligned} & I=\int \frac{\sin x}{3+4 \cos ^2 x} d x \\ & \text { Put } \cos x=t \Rightarrow-\sin x d x=d t \\ & =\frac{-1}{4} \int \frac{d t}{t^2+\binom{\sqrt{3}}{2}^2} \\ & I=-\frac{1}{\text { 4. } \frac{\sqrt{3}}{2}} \cdot \tan ^{-1} \frac{t}{\left(\frac{\sqrt{3}}{2}\right)}+c=\frac{-1}{2 \sqrt{3}} \tan ^{-1} \frac{2 t}{\sqrt{3}}+c \\ & \Rightarrow I=\frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+c \\ & \end{aligned}$

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