Math, asked by Ujagar2480, 9 months ago

Evaluate integral cosec2x.dx

Answers

Answered by Anonymous
4

Answer:

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) + sin 2x/(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

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