Question

The value of $\int (1+tanxtan\frac{x}{2})dx$ is equal to

• A. $\int (\tan x\cot \frac{x}{2}-1)dx$
• B. $\int \sec xdx$
• C. $ln|\sec x-\tan x|+C$
• D. $ln\left|\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right|+C$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $\int (\tan x\cot \frac{x}{2}-1)dx$
B $\int \sec xdx$
C $ln|\sec x-\tan x|+C$
D $ln\left|\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}\right|+C$

$\int 1+\tan \frac{x}{2}\tan xdx=\int \frac{\cos \frac{x}{2}\cos x+\sin \frac{x}{2}\sin x}{\cos x\cos \frac{x}{2}}dx$

$=\int \frac{\cos \frac{x}{2}}{\cos x\cos \frac{x}{2}}dx$ ............ Using, $cos(a-b)=\cos a\cdot \cos b+\sin a\cdot \sin b$

$=\int \sec xdx$

$=\ln |\sec x-\tan x|+c$

Now $\int (\tan x\cot \frac{x}{2}-1)=\int \frac{\sin \frac{x}{2}dx}{\cos x\sin \frac{x}{2}}=\int \sec xdx=\ln |\sec x-\tan x|$

$=\ln \left|\tan (\frac{\pi }{4}+\frac{x}{2})\right|=\ln \left|\frac{1-\tan \frac{x}{2}}{1-\cos \frac{x}{2}}\right|+c$