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Please solve RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 9 maths textbook solution

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Answer: 2 \sqrt{x-\cos x}+c

Hint: Use substitution method to solve this integral.

Given: \int \frac{1+\sin x}{\sqrt{x-\cos x}} d x


        \text { Let } I=\int \frac{1+\sin x}{\sqrt{x-\cos x}} d x

        \begin{aligned} &\text { Put } x-\cos x=t \Rightarrow(1+\sin x) d x=d t \\ &\Rightarrow(1+\sin x) d x=d t \text { then } \end{aligned}

        \begin{aligned} I &=\int \frac{1}{\sqrt{t}} d t=\int t^{\frac{-1}{2}} d t \\ &=\frac{t^{\frac{-1}{2}+1}}{\frac{-1}{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] \end{aligned}

            =2 \sqrt{t}+c

            =2 \sqrt{x-\cos x}+c\; \; \; \; \; [\because t=x-\cos x]

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