Question

Q1PLEASE SOLVE TTHIs



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Answer 1

Step-by-step explanation:

Answer: π²/ab

$\begin{gathered}2I=\int\limits^\pi_0{\frac{\pi.dx}{a^2\cos^2x+b^2\sin^2x}}\\\;\\2I=\pi\int\limits^\pi_0{\frac{dx}{a^2\cos^2x+b^2\sin^2x}}\\\;\\2I=\pi\int\limits^\pi_0{\frac{dx}{\cos^2x(a^2+b^2\tan^2x)}}\\\;\\2I=\pi\int\limits^\pi_0{\frac{\sec^2x.dx}{a^2+b^2\tan^2x}} [tex]2I=\frac{\pi}{b^2}\int\limits^\pi_0{\frac{\sec^2x.dx}{\frac{a^2}{b^2}+\tan^2x}}\\\;\\2I=\frac{2\pi}{b^2}\int\limits^\frac{\pi}{2}_0{\frac{\sec^2x.dx}{\frac{a^2}{b^2}+\tan^2x}}\;\;\;\;\;\;\;(Using\;property\;VI)\\\;\\I=\frac{\pi}{b^2}\int\limits^\frac{\pi}{2}_0{\frac{\sec^2x.dx}{\frac{a^2}{b^2}+\tan^2x}}\\\;\\Let,\\\;\\\tan x=t\implies \sec^2xdx=dt\\\;\\When,\;\;x=0\;\;;t=0\\\;\\x=\frac{\pi}{2}\;\;,\;\;t=\infty\\\;\\I=\frac{\pi}{b^2}\int\limits^\infty_0{\frac{dt}{\frac{a^2}{b^2}+t^2}}\end{gathered}$

$}(\infty)-\tan^{-1}(0)]\\\;\\I=\frac{\pi}{ab}[\frac{\pi}{2}-0]\\\;\\I=\frac{\pi^2}{ab}\end{gathered}</p><p></p><p>Note:-</p><p>$

$\begin{gathered}1.\;\int{\frac{dx}{a^2+x^2}}=\frac{1}{a}\tan^{-1}(\frac{x}{a})\\\;\\2\;\int\limits^{2a}_0{f(x)dx}=2\int\limits^a_0{f(x)dx},\;\text{if}\;\;\;f(2a-x)=f(x)\end{gathered}$