Question

PLEASE SOLVE THIS......​

Answer 1

Answer:  π²/ab

Step-by-step explanation:

Let,

$I=\int\limits^\pi_0{\frac{x}{a^2\cos^2x+b^2\sin^2x}}\,dx\;\;\;..........i)\\\;\\I==\int\limits^\pi_0{\frac{\pi-x}{a^2\cos^2(\pi-x)+b^2\sin^2(\pi-x)}}\,dx\\\;\\I=\int\limits^\pi_0{\frac{\pi-x}{a^2\cos^2x+b^2\sin^2x}}\,dx\;\;\;\;......ii)\\\;\\\text{Adding equation i) and ii)}\\\;\\2I=\int\limits^\pi_0{\frac{x.dx}{a^2\cos^2x+b^2\sin^2x}}+\int\limits^\pi_0{\frac{(\pi-x).dx}{a^2\cos^2x+b^2\sin^2x}}$$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}}</p><p>[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}}$

$I=\frac{\pi}{b^2}[\frac{b}{a}.\tan^{-1}(\frac{bt}{a})]\limits^{\infty}_0\\\;\\I=\frac{\pi}{ab}[\tan^{-1}(\infty)-\tan^{-1}(0)]\\\;\\I=\frac{\pi}{ab}[\frac{\pi}{2}-0]\\\;\\I=\frac{\pi^2}{ab}$

Note:-

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