Q105 / 180 · COMEDK 2019 · Mathematics

$\int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} d x=$

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

Solution

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

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

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