Question

$\int \frac{dx}{a^2{\cos }^{2}x+b^2{\sin }^{2}x}$ is equal to

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

Answer 1

Answer: (C)

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

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

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

$I=\int \frac{se{c}^{2}xdx}{a^2(batnx{)}^2}$

Let t= b tan x

$\frac{dt}{b}=se{c}^{2}xdx$

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

$I=\frac{1}{b}\left[\frac{1}{a}ta{n}^{-1}\right(\frac{t}{a}\left)\right]+c$

$I=\frac{1}{ab}ta{n}^{-1}\left(\frac{btanx}{a}\right)+c$