The value of $\int_{0}^{\pi/4} \frac{dx}{a^2 cos^2 x + b^2 sin^2 x} $ is

Q: The value of $\displaystyle \int_{0}^{\pi/4} \frac{dx}{a^2 cos^2 x + b^2 sin^2 x} $ is

(A) $\displaystyle \frac{1}{ab} tan^{-1} (\frac{b}{a}) ; a > 0 , b > 0$

(B) $\displaystyle \frac{1}{ab} tan^{-1} (\frac{b}{a}) ; a < 0 , b < 0$

(D) $\displaystyle \frac{1}{ab} tan^{-1} (\frac{a}{b}) + \frac{1}{ab}$

Ans: (A) , (B) , (C)

Sol: $\displaystyle I = \int_{0}^{\pi/4} \frac{dx}{a^2 cos^2 x + b^2 sin^2 x} $

Dividing N^r and D^r by cos²x ,

$\displaystyle I = \int_{0}^{\pi/4} \frac{sec^2 x dx}{a^2 + b^2 tan^2 x} $

Put tanx = t ; sec²x dx = dt

$\displaystyle I = \int_{0}^{1} \frac{dt}{a^2 + b^2 t^2} $

$\displaystyle I = \frac{1}{b^2} \int_{0}^{1} \frac{dt}{t^2 + \frac{a^2}{b^2}} $

$\displaystyle I = \frac{1}{b^2} \int_{0}^{1} \frac{dt}{t^2 + (\frac{a}{b})^2} $

$\displaystyle = \frac{b}{a b^2} [tan^{-1}\frac{b t}{a}]_{0}^{1} $

$\displaystyle = \frac{1}{a b} [tan^{-1}\frac{b}{a}] $ ; (a ≠0 , b≠ 0)

Also for a = 1, b = 1

$\displaystyle I = \int_{0}^{\pi/4}dx = \frac{\pi}{4} $

Hence (A), (B), (C) are correct answers.