Question

$\int \frac{dx}{5+4cosx}=ata{n}^{-1}\left(btan\frac{x}{2}\right)+c$, then -

• A. $a=\frac{2}{3}$, $b=-\frac{1}{3}$
• B. $a=\frac{2}{3}$, $b=\frac{1}{3}$
• C. $a=-\frac{2}{3}$, $b=\frac{1}{3}$
• D. $a=-\frac{2}{3}$, $b=-\frac{1}{3}$

Answer 1

The correct option is B $a=\frac{2}{3}$, $b=\frac{1}{3}$

Let $I=\int \frac{1}{5+4\cos x}dx$

Substitute $t=\tan \frac{x}{2}\Rightarrow dt=\frac{1}{2}{\sec }^2\frac{x}{2}dx=\frac{1}{2}(1+{\tan }^2\frac{x}{2})dx⟹dx=\frac{2}{t^2+1}$
$I=\int \frac{2}{(t^2+1)(\frac{4(1-t^2)}{t^2+1}+5)}dt$
$=\int \frac{2}{t^2+9}dt$
$=\frac{2}{3}{\tan }^{-1}\left(\frac{t}{3}\right)$ ......... $\because \int \frac{1}{a^2+x^2}dx=\frac{1}{a}{\tan }^{-1}\left(\frac{x}{a}\right)+c$

$=\frac{2}{3}{\tan }^{-1}\left(\frac{\tan \frac{x}{2}}{3}\right)$

Comparing with

$ata{n}^{-1}\left(btan\frac{x}{2}\right)+c$

We get $a=\frac{2}{3}$, $b=\frac{1}{3}$