Question

${\int }_0^{\pi /2}\frac{dx}{(a^{2}co{s}^{2}x+b^{2}si{n}^{2}x{)}^2}$

Answer 1

${\int }_0^{\pi /2}\frac{dx}{(a^{2}co{s}^{2}x+b^{2}si{n}^{2}x{)}^2}LetI={\int }_0^{\pi /2}\frac{dx}{(a^{2}co{s}^{2}x+b^{2}si{n}^{2}x{)}^2}\text{Divide~ numerator~ and ~denominator~ by~we~ get}co{s}^{4}xI={\int }_0^{\pi /2}\frac{se{c}^{4}xdx}{(a^2+b^{2}ta{n}^{2}x{)}^2}={\int }_0^{\pi /2}\frac{(1+ta{n}^{2}x)se{c}^{2}xdx}{(a^2+b^{2}ta{n}^{2}x{)}^2}Puttanx=t\Rightarrow se{c}^{2}xdx=dtAsx\to 0,thent\to 0andx\to \frac{\pi }{2},thent\to \infty I={\int }_0^{\infty }\frac{(1+t^2)}{(a^2+b^2{t}^2{)}^2}Now,\frac{1+t^2}{(a^2+b^2{t}^2{)}^2}[let{t}^2=u]\frac{1+u}{(a^2+b^{2}u{)}^2}=\frac{A}{(a^2+b^{2}u)}+\frac{B}{(a^2+b^{2}u{)}^2}\Rightarrow 1+u=A(a^2+b^{2}u)+B\text{on comparing the copefficient bt x and constant term on both sides we get}{a}^{2}A+B=1and{b}^{2}A=1\therefore A=\frac{1}{b^2}Now,\frac{a^2}{b^2}+B=1\Rightarrow B=1-\frac{a^2}{b^2}=\frac{b^2-a^2}{b^2}\therefore I={\int }_0^{\infty }\frac{(1+t^2)}{(a^2+b^2{t}^2{)}^2}=\frac{1}{b^2}{\int }_0^{\infty }\frac{dt}{a^2+b^2{t}^2+\frac{b^2-a^2}{b^2}}{\int }_0^{\infty }\frac{dt}{(a^2+b^2{t}^2{)}^2}=\frac{1}{b^2}{\int }_0^{\infty }\frac{dt}{b^2(\frac{a^2}{b^2}+t^2)}+\frac{b^2-a^2}{b^2}{\int }_0^{\infty }\frac{dt}{(a^2+b^2{t}^2{)}^2}=\frac{1}{a{b}^3}{\left[ta{n}^{-1}\left(\frac{tb}{a}\right)\right]}_0^{\infty }+\frac{b^2-a^2}{b^2}(\frac{\pi }{4}.\frac{1}{a^{3}b})=\frac{1}{a{b}^3}[ta{n}^{-1}\infty -ta{n}^{-1}0]+\frac{\pi }{4}.\frac{b^2-a^2}{\left(a^3{b}^3\right)}=\frac{\pi }{2a{b}^3}+\frac{\pi }{4}.\frac{b^2-a^2}{\left(a^3{b}^3\right)}=\pi \left(\frac{2a^2+b^2-a^2}{4a^3{b}^3}\right)=\frac{\pi }{4}\left(\frac{a^2+b^2}{a^3{b}^3}\right)$