Question

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

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

Answer 1

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

$\int \frac{dx}{a^{2}si{n}^{2}x+b^{2}co{s}^{2}x}$

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

$=\int \frac{dx}{co{s}^{2}x(a^{2}ta{n}^{2}x+b^2)}$

$=\int \frac{se{c}^{2}x.dx}{a^{2}ta{n}^{2}x+b^2}$

Let
$\tan x=t$ $\Rightarrow se{c}^{2}x.dx=dt$

Then,
$\int \frac{dt}{a^2{t}^2+b^2}=\frac{1}{b^2}\int \frac{dt}{\frac{a^2{t}^2}{b^2}+1}$

$=\frac{1}{b^2}\int \frac{dt}{(\frac{at}{b}{)}^2+1}$

$=\frac{1}{b^2}\frac{b}{a}{\tan }^{-1}\left(\frac{at}{b}\right)+c$

$=\frac{1}{ab}{\tan }^{-1}\left(\frac{at}{b}\right)+c$

$=\frac{1}{ab}{\tan }^{-1}\left(\frac{a\tan x}{b}\right)+c$