Question

Solve :

$\int \frac{1}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx.$

• A. ${\tan }^{-1}\left(\frac{b}{a}\tan x\right)$
• B. $\frac{1}{ab}{\tan }^{-1}\left(\frac{a}{b}\tan x\right)$
• C. $\frac{1}{ab}\tan \left(\frac{b}{a}\tan x\right)$
• D. $\frac{1}{ab}{\tan }^{-1}\left(\frac{b}{a}\tan x\right)$

Answer 1

The correct option is D $\frac{1}{ab}{\tan }^{-1}\left(\frac{b}{a}\tan x\right)$

Let $I=\int \frac{1}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}dx$

Dividing numerator and denominator by ${\cos }^{2}x$, we get

$I=\int \frac{\frac{1}{{\cos }^{2}x}}{a^2+b^2\frac{{\sin }^{2}x}{{\cos }^{2}x}}dx$

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

Put $b\tan x=t\Rightarrow b{\sec }^{2}xdx=dt$

$I=\frac{1}{b}\int \frac{dt}{a^2+t^2}=\frac{1}{b}\frac{1}{a}{\tan }^{-1}\frac{t}{a}=\frac{1}{ab}{\tan }^{-1}\left(\frac{b}{a}\tan x\right)$