Browse Questions

# Evaluate the following:$\int\limits_0^{\Large\frac{\pi}{2}}\large\frac{dx}{(a^2\cos^2x+b^2\sin^2x)^2}\quad\normalsize \text{(Hint:Divide Numerator and Denominator by$\cos^4x)$}$

Toolbox:
• $\large\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}\big(\frac{x}{a}\big)+c.$
• If f(x) is substituted by $f(t)$,then $f'(x)=f'(t).$Hence $\int f'(x) dx=\int f'(t)dt.$
• $\tan^{-1}\infty=\large\frac{\pi}{2}$
Step 1:
Let $I=\int_0^{\large\frac{\pi}{2}}\large\frac{dx}{(a^2\cos^2 x+b^2\sin^2x)^2}$
Divide the numerator and denominator by $\cos^4x$
$I=\int_0^{\large\frac{\pi}{2}}\large\frac{\Large\frac{1}{\cos^4x}}{(a^2\Large\frac{\cos^2 x}{cos^2x}+b^2\Large\frac{\sin^2x}{\cos^2x})^2}dx.$
$\;\;=\int_0^{\large\frac{\pi}{2}}\large\frac{\sec^4xdx}{a^2+b^2\cos^2x)^2}=\int_0^{\large\frac{\pi}{2}}\large\frac{\sec^2x.\sec^2x dx}{(a^2+b^2\tan^2x)^2}$
But $\sec^2x=1+\tan^2x$
$\;\;=\int_0^{\large\frac{\pi}{2}}\large\frac{\sec^2x.(1+\tan^2x) dx}{(a^2+b^2\tan^2x)^2}$
Put $b\tan x=a\tan\theta$,on differentiating we get
$b\sec^2xdx=a\sec^2\theta d\theta=\large\frac{a}{b}\normalsize sec^2\theta d\theta$
Now when we substitute for $b\tan x$,the limits also change.
When $x=0$,$\theta=0$,when $x=\large\frac{\pi}{2},\theta=\large\frac{\pi}{2}$
On substituting we get,
$I=\int_0^{\large\frac{\pi}{2}}\large\frac{(1+\large\frac{a^2}{b^2}\tan^2\theta).\frac{a}{b}\sec^2\theta d\theta}{(a^2+a^2\tan^2\theta)^2}$
$\;\;=\int_0^{\large\frac{\pi}{2}}\large\frac{(b^2+a^2\tan^2\theta).a\sec^2\theta d\theta}{b^3a^4\sec^4\theta}$
Step 2:
On simplifying we get,
$\;\;=\int_0^{\large\frac{\pi}{2}}\large\frac{1}{a^3b^3}\normalsize (b^2+a^2\tan ^2\theta).\cos^2\theta d \theta$
$\;\;=\large\frac{1}{a^3b^3}\int_0^{\large\frac{\pi}{2}}\normalsize (b^2\cos ^2\theta+a^2\sin^2\theta)d \theta$
On splitting the terms we get,
$\;\;=\large\frac{1}{a^3b^3}\int_0^{\large\frac{\pi}{2}}\normalsize b^2\cos ^2\theta d\theta +\large\frac{1}{a^3b^3}\int_0^{\large\frac{\pi}{2}}\normalsize a^2\sin ^2\theta d\theta$
$\cos^2\theta=\large\frac{1+\cos 2\theta}{2}$ and $\sin^2\theta=\large\frac{1-\cos^2\theta}{2}$
$\;\;=\large\frac{1}{a^3b^3}\int_0^{\large\frac{\pi}{2}}\frac{ b^2(1+\cos 2\theta) d\theta}{2} +\large\frac{1}{a^3b^3}\int_0^{\large\frac{\pi}{2}}\frac {a^2(1-\cos 2\theta )}{2}\normalsize d\theta$
On integrating we get,
$\;\;=\large\frac{1}{a^3b^3}\begin{bmatrix}\frac{b^2}{2}\normalsize(\theta+\frac{1}{2}\sin2\theta)\end{bmatrix}_0^{\large\frac{\pi}{2}}+\frac{a^2}{2}\begin{bmatrix}\normalsize (\theta-\frac{1}{2}\sin 2\theta)\end{bmatrix}_0^{\large\frac{\pi}{2}}$
On applying limits,
$\;\;=\frac{1}{a^3b^3}\begin{bmatrix}\frac{b^2}{2}(\frac{\pi}{2}+\frac{1}{2}\sin 2.\frac{\pi}{2})\end{bmatrix}-[0+\frac{1}{2}\sin 0]+\frac{a^2}{2}\begin{bmatrix}\frac{\pi}{2}-\frac{1}{2}\sin 2.\frac{\pi}{2}-0+\frac{1}{2}\sin 0\end{bmatrix}$
$\;\;=\large\frac{1}{a^3b^3}[\normalsize b^2.\frac{\pi}{4}+\frac{\pi a^2}{4}]$
$\;\;= \large\frac{\pi}{4}\frac{(a^2+b^2)}{a^3b^3}$