KCET · Maths · Indefinite Integration
\(\int \sqrt{\operatorname{cosec} x-\sin x} d x\) is equals to
- A \(\frac{\sqrt{\sin x}}{2}+C\)
- B \(2 \sqrt{\sin x}+C\)
- C \(\frac{2}{\sqrt{\sin x}}+C\)
- D \(\sqrt{\sin x}+C\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{\sin x}+C\)
Step-by-step Solution
Detailed explanation
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 \(\ddot{i}-k\)
\(\cos x d x=d k\)
\(\begin{aligned} & I=\int \frac{d k}{\sqrt{k}}=\frac{k^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+C \\ & =\frac{k^{\frac{1}{2}}}{\frac{1}{2}}+C=2 \sqrt{k}+C \\ & I=2 \sqrt{\sin x}+C \quad \quad[\because k=\sin x]\end{aligned}\)
\(\begin{aligned} I & =\int \frac{\sqrt{1-\sin ^2 x}}{\sqrt{\sin x}} d x \\ & =\int \frac{\cos x}{\sqrt{\sin x}} d x\end{aligned}\)
Ful, sul \(\ddot{i}-k\)
\(\cos x d x=d k\)
\(\begin{aligned} & I=\int \frac{d k}{\sqrt{k}}=\frac{k^{\frac{-1}{2}+1}}{\frac{-1}{2}+1}+C \\ & =\frac{k^{\frac{1}{2}}}{\frac{1}{2}}+C=2 \sqrt{k}+C \\ & I=2 \sqrt{\sin x}+C \quad \quad[\because k=\sin x]\end{aligned}\)
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