Question

The value of the integral $I={\int }_0^{\pi /4}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$ is

• A. $\frac{1}{ab}{\tan }^{-1}\frac{b}{a}(a>0,b>0)$
• B. $\frac{1}{ab}{\tan }^{-1}\frac{b}{a}(a<0,b<0)$
• C. $\frac{\pi }{4}(a=1,b=1)$
• D. $\frac{1}{ab}{\tan }^{-1}\frac{a}{b}+\frac{1}{ab}$

Answer 1

Answer: (A), (B), (C)

$\frac{1}{ab}{\tan }^{-1}\frac{b}{a}(a>0,b>0)$
$\frac{\pi }{4}(a=1,b=1)$
$\frac{1}{ab}{\tan }^{-1}\frac{b}{a}(a<0,b<0)$

$I={\int }_0^{\pi /4}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

$={\int }_0^{\pi /4}1dx=\frac{\pi }{4}$

$[a=1,b=1]$ Also,

$I={\int }_0^{\pi /4}\frac{{\sec }^{2}x}{a^2{b}^2{\tan }^{2}x}dx={\int }_0^1\frac{dt}{a^2+b^2{t}^2}$ $[t=\tan x]$

$\frac{1}{b^2}={\int }_0^1\frac{dx}{t^2+\left(a^2/b^2\right)}=\frac{b}{a{b}^2}{\tan }^{-1}{\frac{bt}{a}|}_0^1$ $[a\ne 0,b\ne 0]$

$=\frac{1}{ab}{\tan }^{-1}\frac{b}{a}$