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need solution for RD Sharma maths class 12 chapter Indefinite Integrals exercise 18.9 question 47

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

Hint: Use substitution method to solve this integral.

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


        \begin{aligned} &\text { Let } I=\int \frac{\sin \sqrt{x}}{\sqrt{x}} d x \\ &\text { Put } \sqrt{x}=t \Rightarrow \frac{1}{2 \sqrt{x}}\; d x=d t \\ &\Rightarrow d x=2 \sqrt{x} \; d t \text { then } \end{aligned}

        I=\int \frac{\sin t}{\sqrt{x}} 2 \sqrt{x} d t=2 \int \sin t\; d t

           =2[-\cos t]+\mathrm{c} \quad\left[\because \int \sin x \; d x=-\cos x+c\right]

           \begin{aligned} &=-2 \cos t+c \\ &=-2 \cos \sqrt{x}+c \quad[\because t=\sqrt{x}] \end{aligned}

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