# What is the integral of cosec 2x dx ?

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### 2 Answers

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**

We'll write the function cosec 2x = 1/sin 2x

We'll use the identity of double angle:

sin 2x = 2sin x*cos x

Int cosec 2x dx =Int dx/sin 2x

Int dx/sin 2x = Int dx/2sinx*cosx

We'll use Pythagorean identity:

(sin x)^2 + (cos x)^2 = 1

Int dx/2sinx*cosx = Int [(sin x)^2 + (cos x)^2]dx/2sinx*cosx

Int dx/2sinx*cosx =(1/2)*[Int sin xdx/cos x + Int cos xdx/sin x]

(1/2)*[Int sin xdx/cos x + Int cos xdx/sin x] = (1/2)[-ln(cosx) + ln(sin x)]

**Int cosec 2x dx = -ln(sqrt (cos x)) + ln (sqrt(sin x)) + C**