Q108 / 180 · KCET 2023 · Mathematics

$\int \sqrt{\operatorname{cosec} x-\sin x} d x$ is equals to

  • A
    $\frac{\sqrt{\sin x}}{2}+C$
  • Correct answer
    B
    $2 \sqrt{\sin x}+C$
  • C
    $\frac{2}{\sqrt{\sin x}}+C$
  • D
    $\sqrt{\sin x}+C$

Solution

Let $I=\int \sqrt{\operatorname{cosec} x-\sin x} d x$ $\begin{aligned} I & =\int \frac{\sqrt{1-\sin ^2 x}}{\sqrt{\sin x}} d x \\ & =\int \frac{\cos x}{\sqrt{\sin x}} d x\end{aligned}$ Ful, sul $\ddot{i}-k$ $\cos x d x=d k$ $\begin{aligned} & I=\int \frac{d k}{\sqrt{k}}=\frac{k^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+C \\ & =\frac{k^{\frac{1}{2}}}{\frac{1}{2}}+C=2 \sqrt{k}+C \\ & I=2 \sqrt{\sin x}+C \quad \quad[\because k=\sin x]\end{aligned}$

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

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