Question

Prove that ${\int }_0^{\frac{\pi }{2}}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}=\frac{\pi }{2ab}(a,b>0)$.

Answer 1

Let $LHS:I={\int }_0^{\frac{\pi }{2}}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

Divide and multiply by ${\cos }^{2}x$.

$I={\int }_0^{\frac{\pi }{2}}\frac{{\sec }^{2}xdx}{a^2+b^2{\tan }^{2}x}$

Put $\tan x=t\Rightarrow {\sec }^{2}xdx=dt$, and limits will be

Lower limit $x=0\Rightarrow t=0$

Upper limit $x=\frac{\pi }{2}\Rightarrow t=\infty$

$I={\int }_0^{\infty }\frac{dt}{a^2+b^2{t}^2}=\frac{1}{b^2}{\int }_0^{\infty }\frac{dt}{{\left(\frac{a}{b}\right)}^2+t^2}=\frac{1}{b^2}\left(\frac{b}{a}\right){\left|{\tan }^{-1}\right(\frac{at}{b}\left)\right|}_0^{\frac{\infty }{2}}$

$=\frac{1}{ab}|\frac{\pi }{2}-0|=\frac{\pi }{2ab}=RHS$

Hence proved.