Q114 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

$\int \frac{d x}{9 \cos ^2 2 x+16 \sin ^2 2 x}=$

  • A
    $\frac{1}{25} \tan ^{-1}\left(\frac{3}{4} \sec ^2 2 x\right)+c$
  • B
    $\frac{1}{25} \tan ^{-1}\left(\frac{4}{3} \sec ^2 2 x\right)+c$
  • C
    $\frac{1}{24} \tan ^{-1}\left(\frac{3}{4} \tan 2 x\right)+c$
  • Correct answer
    D
    $\frac{1}{24} \tan ^{-1}\left(\frac{4}{3} \tan 2 x\right)+c$

Solution

$I=\int \frac{1 d x}{9 \cos ^2 2 x+16 \sin ^2 2 x}$ $\begin{aligned} & =\int \frac{\sec ^2 2 x d x}{9+16 \tan ^2 2 x}=\frac{1}{9} \int \frac{\sec ^2 2 x d x}{\left[1+\left(\frac{4}{3} \tan 2 x\right)^2\right]} \\ & =\frac{1}{24} \tan ^{-1}\left(\frac{4}{3} \tan 2 x\right)+C .\end{aligned}$

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

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