Question

integration of sin2x/(1-cos2x) (2-cos2x)

Answer 1

hint only:- put cos2x = t u get sin2x *2 after this use partial fraction method for solving this problem it become very simple

Answer 2

Answer:

$\int \frac{sin \: 2x}{(1 - cos \: 2x)(2 - cos \: 2x)} dx\\\\ =log \bigg( \sqrt{ \bigg| \frac{1 - cos \: 2x}{2 - cos \: 2x} \bigg| } \bigg) + C$

Step-by-step explanation:

To integrate

$\int \frac{sin \: 2x}{(1 - cos \: 2x)(2 - cos \: 2x)} dx \\ \\ let \: cos \: 2x = t \\ \\ - 2 \: sin \: 2x \: dx = dt \\ \\ sin \: 2x \: dx = \frac{dt}{ - 2} \\ \\ so \\ \\ - \frac{1}{2} \int \frac{dt}{(1 - t)(2 - t)}$

Now solve this by partial fraction

$\frac{1}{(1 - t)(2 - t)} = \frac{A}{1 - t} + \frac{B}{2 - t} \\ \\ 1 = A(2 - t) + B(1 - t) \\ \\ 1 = 2A + B + t( - A - B) \\ \\ 2A + B= 1 \\ - A- B = 0 \\ \\ add \: both \: equations \\ \\ A = 1 \\ \\ B = - 1$

$- \frac{1}{2} \int \frac{1}{1 - t}dt + \frac{1}{2} \int \frac{1}{2 - t}dt \\ \\ = \frac{1}{2} . log( |1 - t| ) - \frac{1}{2} log( |2 - t| ) + C \\ \\ = log \bigg( \sqrt{ \bigg| \frac{1 - t}{2 - t} \bigg| } \bigg) + C$

Undo substitution

$\int \frac{sin \: 2x}{(1 - cos \: 2x)(2 - cos \: 2x)} dx= log \bigg( \sqrt{ \bigg| \frac{1 - cos \: 2x}{2 - cos \: 2x} \bigg| } \bigg) + c$

Hope it helps you.