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need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 11

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Answer:\frac{-2}{\sqrt{\sin x}}+c

Hint: Use substitution method to solve this integral.

Given: \int \frac{\cot x}{\sqrt{\sin x}} d x


        \begin{aligned} &\text { Let } I=\int \frac{\cot x}{\sqrt{\sin x}} d x \\ &\text { Put } \sin x=t \Rightarrow \cos x d x=d t \\ &\Rightarrow d x=\frac{d t}{\cos x} \text { then } \end{aligned}

        I=\int \frac{\cot x}{\sqrt{t}} \frac{d t}{\cos x}=\int \frac{\cos x}{\sin x} \cdot \frac{1}{\sqrt{t}} \cdot \frac{d t}{\cos x} d t \quad\left[\because \cot x=\frac{\cos x}{\sin x}\right]

           \begin{aligned} &=\int \frac{1}{t . t^{\frac{1}{2}}} d t \quad[\because t=\sin x] \\ &=\int \frac{1}{t^{1+\frac{1}{2}}} d t=\int \frac{1}{t^{\frac{3}{2}}} \mathrm{dt} \end{aligned}

       I=\int t^{\frac{-3}{2}} d t

           \begin{aligned} &=\frac{t^{\frac{-3}{2}+1}}{\frac{-3}{2}+1}+c=\frac{t^{\frac{-1}{2}}}{\frac{-1}{2}}+\mathrm{c} \quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right] \\ &=-2 \frac{1}{\sqrt{t}}+c=\frac{-2}{\sqrt{\sin x}}+c\; \; \; \; [\because t=\sin x] \end{aligned}

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