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#### Please Solve R.D.Sharma class 12 Chapter 19 Definite Integrals Exercise 19.1  Question 6 Maths textbook Solution.

Answer:     $\frac{\pi }{2ab}$

Hint:Use indefinite formula to solve the integral and then put the value of limit to get the required answer

Given:$\int_{0}^{\infty }\frac{1}{a^{2}+b^{2}x^{2}}dx$

Solution:

$\int_{0}^{\infty }\frac{1}{a^{2}+b^{2}x^{2}}dx$

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Answer:$\frac{\pi }{2ab}$

Hint:Use indefinite formula to solve the integral and then put the value of limit to get the required answer

Given:

$\int_{0}^{\infty }\frac{1}{a^{2}+b^{2}x^{2}}dx$

Solution:

$\int_{0}^{\infty }\frac{1}{a^{2}+b^{2}x^{2}}dx$

\begin{aligned} &=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{\frac{a^{2}}{b^{2}}+x^{2}} d x \\ &=\frac{1}{b^{2}} \int_{0}^{\infty} \frac{1}{x^{2}+\left(\frac{a}{b}\right)^{2}} d x \end{aligned}

$=\frac{1}{b^{2}}\left[\frac{1}{\frac{a}{b}} \tan ^{-1}\left(\frac{x}{\frac{a}{b}}\right)\right]_{0}^{\infty} \quad\left[\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)\right]$

\begin{aligned} &=\frac{1}{b^{2}} \frac{b}{a}\left[\tan ^{-1}\left(\frac{b x}{a}\right)\right]_{0}^{\infty}\\ &=\frac{1}{a b}\left[\tan ^{-1} \infty-\tan ^{-1} 0\right] \quad\left[\begin{array}{l} \tan \frac{\pi}{2}=\infty \Rightarrow \tan ^{-1} \infty=\frac{\pi}{2} \\ \tan 0=0 \Rightarrow \tan ^{-1} 0=0 \end{array}\right]\\ &=\frac{1}{a b}\left[\frac{\pi}{2}-0\right] \end{aligned}

$=\frac{\pi }{2ab}$