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Explain solution RD Sharma class 12 chapter 19 Definite Integrals exercise Very short answer type question 12 maths

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Answer: \frac{\pi }{12}

Hint: You must know the integration rules of trigonometric function with its limits

Given: \int_{0}^{3} \frac{1}{x^{2}+9} d x

Solution:  \int_{0}^{3} \frac{1}{x^{2}+9} d x

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

\begin{aligned} &=\frac{1}{3}\left[\tan ^{-1} \frac{x}{3}\right]_{0}^{3} \\\\ &=\frac{1}{3}\left[\tan ^{-1} 1-\tan ^{-1} 0\right] \end{aligned}

\begin{aligned} &=\frac{1}{3}\left[\frac{\pi}{4}\right] \\\\ &=\frac{\pi}{12} \end{aligned}

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