Question

Integral Cosec2x dx ?

Answer 1

Answer:

integral cosec(ax+b)=1/a log|cosec ax - cot ax|+C

therefore

integral cosec 2x dx=

1/2 log|cosec 2x - cot 2x|+C

Answer 2

Step-by-step explanation:

We need to find the integral of cosec 2x dx.

Int[ cosec 2x dx]

=> Int[ 1/ sin 2x dx]

=> Int[ (sin 2x)/ (sin 2x)^2 dx]

=> Int[ (sin 2x)/(1 - (cos 2x)^2) dx]

=> Int[ (sin 2x)/(1 - cos 2x)(1 + cos 2x) dx]

=> 1/2(Int [ sin 2x / (1 - cos 2x) + sin2x/(1 + cos 2x) dx]

=> 1/4[ ln| 1 - cos 2x | - ln|( 1 + cos 2x)]

=> 1/4[ - ln(|1 + cos 2x | / | 1 - cos 2x|) ]

=> 1/4[ - ln((1 + cos 2x)^2 / ((1 - cos 2x)^2) ]

=> 1/4[ - ln((1 + cos 2x)^2 / (sin 2x)^2]

=> 1/2[ - ln|(1/(sin 2x) + (cos 2x) / (sin 2x)|]

=> 1/2[ - ln|(cosec 2x) + (cot 2x)|]

=> -1/2( ln | cosec 2x + cot 2x| + C

The required integral is -1/2(ln | cosec 2x + cot 2x| + C