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explain solution RD Sharma class 12 chapter Indefinite Integrals exercise 18.9 question 52 maths

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Answer: -\cos \left(\tan ^{-1} x\right)+c

Hint: Use substitution method to solve this integral.

Given:   \int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}} d x


        \text { Let } I=\int \frac{\sin \left(\tan ^{-1} x\right)}{1+x^{2}} d x

        \begin{aligned} \text { put } \tan ^{-1} x=t & \Rightarrow \frac{1}{1+x^{2}} d x=d t \\ & \Rightarrow d x=\left(1+x^{2}\right) d t \end{aligned}

        I=\int \frac{\sin t}{1+x^{2}}\left(1+x^{2}\right) d t=\int \sin t \; d t

           \begin{array}{ll} =-\cos t+c & {\left[\because \int \sin x \; d x=-\cos x+c\right]} \\ =-\cos \left(\tan ^{-1} x\right)+c & {\left[\because t=\tan ^{-1} x\right]} \end{array}

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