Question

Integration of sinx/1-cosx

Answer 1

= integral( sin x dx) + integral (sin x cos x dx) 

For evaluating the second term, 

sin x cos x dx = 1/2 * 2 sin x cos x dx 
= 1/2 sin 2x dx 

let 2x = t. 

Then dt = 2 dx 

So it becomes 

1/4 sin t dt 

Integrating, it becomes, -1/4 cos t = -1/4 cos 2x 

So final answer is -cos x - 1/4 cos 2x

Answer 2

Answer:

$<b>$
I = $\int$( sinx/1-cosx ) dx

Put 1 - cos x = t


Now differentiate both sides wrt. x


sin x dx = dt


sin x dx = dt


Now Integrate:


I = $\int$( dt/t)


I = $\int$ (1.dt/dx)


I = log | t | + c


I = log | 1 + cos x | + c



#Be Brainly.