Question

The value of the integral ${\int }_{-\pi /4}^{\pi /4}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$

• A. $\frac{2}{ab}{\tan }^{-1}\frac{b}{a}(a>0,b>0)$
• B. $\frac{2}{ab}{\tan }^{-1}\frac{b}{a}(a<0,b<0)$
• C. $\frac{\pi }{2}(a=1,b=1)$
• D. $\frac{2}{ab}{\tan }^{-1}\frac{a}{b}+\frac{1}{ab}$

Answer 1

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

The correct options are
A $\frac{2}{ab}{\tan }^{-1}\frac{b}{a}(a>0,b>0)$
B $\frac{2}{ab}{\tan }^{-1}\frac{b}{a}(a<0,b<0)$
C $\frac{\pi }{2}(a=1,b=1)$

Let $I={\int }_{-\pi /4}^{\pi /4}\frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}2{\int }_0^{\frac{\pi }{4}}\frac{dx}{a^2{\cos }^{2}x+b^{2}si{n}^{2}x}$
Multiply numerator and denominator by ${\sec }^{2}x$
$I=2{\int }_0^{\frac{\pi }{4}}\frac{{\sec }^{2}xdx}{a^2+b^2{\tan }^{2}x}$
Put ${\tan }^{2}x=t$, we get
$I=2{\int }_0^1\frac{dt}{a^2+b^2{t}^2}$
$={\left[\frac{2}{ab}{\tan }^{-1}\frac{bt}{a}\right]}_0^1=\frac{2}{ab}{\tan }^{-1}\frac{b}{a}$