Question

Evaluate : ∫1+cosx1−cosxdx

Answer 1

Correct Question:

• $\int \frac{1 + \cos(x) }{1 - \cos(x) }dx$

Answer:

• $\large \bold\red{ - 2 \cot( \frac{x}{2} ) - x + c}$

Step-by-step explanation:

Given,

• $\int \frac{1 + \cos(x) }{1 - \cos(x) }dx$

We know that,

$\bold{1 + \cos(2 \alpha ) = 2 { \cos }^{2} \alpha } \\ and \\ \bold{1 - \cos(2 \alpha ) = 2 { \sin }^{2} \alpha }$

Therefore,

We get,

$= \int \frac{2 { \cos }^{2}( \frac{x}{2}) }{2 { \sin }^{2}( \frac{x}{2} ) } dx \\ \\ = \int { \cot }^{2} ( \frac{x}{2}) dx$

But,

We know that,

$\bold{{ \cot }^{2} \alpha = { \cosec}^{2} \alpha - 1}$

Therefore,

We get,

$= \int( { \cosec }^{2} \frac{x}{2} - 1)dx \\ \\ = \int {cosec}^{2} \frac{x}{2} \: dx - \int \: dx \: \: \: \: \: \: ........(i)$

Now,

Let,

$\frac{x}{2} = t$

On differentiating both sides wrt x,

We get,

$= > dx = 2dt$

Substituting this value in eqn $(i)$,

We get,

$=2\int{cosec}^{2}t-\int\:dx\\\\=-2cot(t)-x+C$

Now,

Putting the value of t,

We get,

$= \large \bold{ - 2 \cot( \frac{x}{2} ) - x + c}$

Where,

C is any arbitrary constant.