Q98 / 180 · KCET 2024 · Mathematics

$\int \frac{\sin x}{3+4 \cos ^2 x} d x$

  • Correct answer
    A
    $-\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+C$
  • B
    $\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\cos x}{3}\right)+C$
  • C
    $\frac{1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{\cos x}{3}\right)+C$
  • D
    $-\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{3}\right)+C$

Solution

$\because I=\int \frac{\sin x}{3+4 \cos ^2 x} d x$ $=\frac{\sin x}{3\left[1+\left(\frac{2 \cos x}{\sqrt{3}}\right)^2\right]} d x$ put $\quad t=\frac{2 \cos x}{\sqrt{3}}$ $d t=\frac{-2 \sin x}{\sqrt{3}} d x$ $I=\int-\frac{\sqrt{3}}{2\left(3 t^2+3\right)} d t$ $=-\frac{1}{2 \sqrt{3}} \int \frac{1}{1+t^2} d t$ $=-\frac{1}{2 \sqrt{3}} \tan ^{-1}(t)+C$ $=\frac{-1}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 \cos x}{\sqrt{3}}\right)+C$

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

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