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Need solution for RD Sharma maths class 12 chapter 19 Definite Integrals exercise Multiple choice question 15

Answers (1)

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Answer:

\frac{\pi}{12}

Given:

\int_{1}^{\sqrt{3}} \frac{1}{1+x^{2}} d x

Hint:

Use formula \int \frac{1}{1+x^{2}} d x=\tan ^{-1} x


Explanation:  

\begin{aligned} I &=\int_{1}^{\sqrt{3}} \frac{1}{1+x^{2}} d x \\\\ &=\left[\tan ^{-1} x\right]_{1}^{\sqrt{3}} \\\\ &=\tan ^{-1} \sqrt{3}-\tan ^{-1} 1 \end{aligned}

    \begin{aligned} &=\frac{\pi}{3}-\frac{\pi}{4} \\\\ &=\frac{4 \pi-3 \pi}{12} \\\\ &=\frac{\pi}{12} \end{aligned}

 

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