show that

$\int\limits_0^{\dfrac \pi2} \dfrac {dx}{(a^2\sin^2x + b^2\cos^2x)}=\dfrac \pi{4ab}\Bigg(\dfrac 1{a^2}+\dfrac 1{b^2}\Bigg)$

Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 8

Year : 2014

Dividing numerator and denominator by $\cos^2 x$ we get

$I=\int\limits_0^{\dfrac \pi2}\dfrac {\sec^2 x\space dx}{a^2\tan^2x+b^2}$

Putting $\tan x=t \hspace {1cm} sec^2 x\space dx=dt $

When $x=0 \space t=0$

$X=\pi/2 \space\space t=\infty $

$I=\int\limits_0^{\infty}\dfrac {dt}{a^2t^2+b^2}$

$=\dfrac 1{a^2}\Bigg[\dfrac 1{\dfrac ba}\tan^{-1}\Bigg(\dfrac t{\dfrac ba}\Bigg)\Bigg]_0^{\infty} $

$=\dfrac 1{ab}\dfrac \pi2$

Now we apply the rule of DUIS

Differentiating both sides w.r. t. a

$\int\limits_0^{\dfrac \pi2}\Bigg[\dfrac{\partial}{\partial a}\Bigg(\dfrac 1{a^2\sin^2 x+b^2\cos^2x}\Bigg)\Bigg]dx=\dfrac \pi{2b}\Bigg(-\dfrac 1{a^2}\Bigg)$

$\therefore \int\limits_0^{\dfrac \pi2}\Bigg[\Bigg(-\dfrac {2a\sin^2x}{a^2\sin^2 x+b^2\cos^2x}\Bigg)\Bigg]dx=\dfrac \pi{2b}\Bigg(-\dfrac 1{a^2}\Bigg)$

$\int\limits_0^{\dfrac \pi2}\Bigg(\dfrac {\sin^2x}{a^2\sin^2 x+b^2\cos^2x}\Bigg)dx=\dfrac \pi{4ab}\Bigg(\dfrac 1{a^2}\Bigg)-----(2)$

Again differentiating both sides w. r. t. b

$\int\limits_0^{\dfrac \pi2}\dfrac {\cos^2x}{a^2\sin^2 x+b^2\cos^2x}dx=\dfrac \pi{4ab}\Bigg(\dfrac 1{a^2}\Bigg)---- (3)$

Adding (2) and (3) we get

$\int\limits_0^{\dfrac \pi2}\dfrac {\sin^2x+\cos^2x}{a^2\sin^2 x+b^2\cos^2x}dx=\dfrac \pi{4ab}\Bigg(\dfrac 1{a^2}+\dfrac 1{b^2}\Bigg)$

$\int\limits_0^{\dfrac \pi2}\dfrac 1{a^2\sin^2 x+b^2\cos^2x}dx=\dfrac \pi{4ab}\Bigg(\dfrac 1{a^2}+\dfrac 1{b^2}\Bigg)$