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Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Multiple choice question 46 maths

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Answer:

1

Hint:

To solve this question we have to use substitution method.

Given:

\int_{0}^{1} \tan \left(\sin ^{-1} x\right) d x

Explanation:

\begin{aligned} &\sin ^{-1} x=t \\\\ &x=\sin t \\\\ &d x=\cos t d t \\\\ &\text { When } x=0, \sin ^{-1} 0=0 \\\\ &\text { When } x=1, \sin ^{-1} 1=\frac{\pi}{2} \end{aligned}

\begin{aligned} &I=\int_{0}^{\frac{\pi}{2}} \tan t \cdot \cos t \; d t \\\\ &\frac{\sin t}{\cos t}=\tan t \\\\ &\sin t=\cos t \cdot \cos t \end{aligned}

\begin{aligned} &=\int_{0}^{\frac{\pi}{2}} \sin t d t \\\\ &I=[-\cos t]_{0}^{\frac{\pi}{2}} \\\\ &I=0-(-1) \\\\ &=1 \end{aligned}

 

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