Question

$A:\int \frac{1}{3+2\cos x}dx=\frac{2}{\sqrt{5}}{\tan }^{-1}(\frac{1}{\sqrt{5}}\tan \frac{x}{2})+c$
$B:$ If $a>b$ then $\int \frac{dx}{a+b\cos x}=\frac{2}{\sqrt{a^2-b^2}}{\tan }^{-1}\left[\sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2}\right]+c$

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true but R is not correct explanation of A
• C. A is true R is false
• D. A is false but R is true.

Answer 1

The correct option is A Both A and R are true and R is the correct explanation of A

$\int \frac{dx}{a+b\cos x}$
$=\int \frac{dx}{a+b\left(\frac{1-{\tan }^{2}x/2}{1+{\tan }^{2}x/2}\right)}$
$=\int \frac{{\sec }^{2}x/2dx}{(a+b)+(a-b){\tan }^{2}x/2}$
$=\int \frac{{\sec }^{2}x/2dx}{(a-b)[\frac{a+b}{a-b}+{\tan }^{2}x/2]}$
Put $\tan \left(x/2\right)=t$
$\frac{1}{2}{\sec }^2\left(x/2\right)dx=dt$
$=\int \frac{2dt}{{(a-b)\left[\right(\sqrt{\frac{a+b}{a-b}})}^2+t^2]}$
$=\frac{2}{(a-b)\sqrt{\frac{a+b}{a-b}}}{\tan }^{-1}\left(\frac{t}{\sqrt{\frac{a+b}{a-b}}}\right)+c$
$\int \frac{dx}{a+b\cos x}=\frac{2}{\sqrt{a^2-b^2}}{\tan }^{-1}\left[\sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2}\right]+c$
$A:\int \frac{1}{3+2\cos x}dx=\frac{2}{\sqrt{5}}{\tan }^{-1}(\frac{1}{\sqrt{5}}\tan \frac{x}{2})+c$
For assertion, put $a=3,b=2(a>b)$
$\int \frac{1}{3+2\cos x}dx=\frac{2}{\sqrt{5}}{\tan }^{-1}(\frac{1}{\sqrt{5}}\tan \frac{x}{2})+c$
Hence, option 'A' is correct.