Question

$\int \frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}(a,b>0)$ is equal to
(where $C$ is integration constant)

• A. $\frac{1}{b^2}{\tan }^{-1}\left(\frac{b}{a}\tan x\right)+C$
• B. $\frac{a}{b}{\tan }^{-1}\left(\frac{b}{a}\tan x\right)+C$
• C. $\frac{1}{ab}{\tan }^{-1}\left(\frac{b}{a}\tan x\right)+C$
• D. $ab{\tan }^{-1}\left(\frac{b}{a}\tan x\right)+C$

Answer 1

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

$I=\int \frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$
Dividing numerator and denominator by ${\cos }^{2}x$
$\Rightarrow I=\int \frac{{\sec }^{2}xdx}{a^2+b^2{\tan }^{2}x}$
Put $\tan x=t$
$\Rightarrow {\sec }^{2}xdx=dt\Rightarrow I=\int \frac{dt}{a^2+b^2{t}^2}\Rightarrow I=\frac{1}{b^2}\int \frac{dt}{t^2+{\left(\frac{a}{b}\right)}^2}\Rightarrow I=\frac{1}{b^2}\cdot \frac{b}{a}{\tan }^{-1}\left(\frac{b}{a}t\right)+C[\because \int \frac{1}{x^2+a^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+C]\therefore I=\frac{1}{ab}{\tan }^{-1}\left(\frac{b}{a}\tan x\right)+C$