Question

Integrate csc^3 2x dx ?

Answer 1

`I=1/16(tan^2(x)+4ln(tan(x))-1/(tan^2(x)))+C`

Explanation:

We want to solve

`I=intcsc^3(2x)dx`

Make a substitution `u=2x=>du=2dx`

`I=1/2intcsc^3(u)du`

Use tangent half-angle substitution `s=tan(u/2)`,
then `csc(u)=(1+s^2)/(2s)` and `du=2/(1+s^2)ds`

`I=1/2int((1+s^2)/(2s))^3 2/(1+s^2)ds`

`color(white)(I)=int(1+s^2)^2/(8s^3)ds`

`color(white)(I)=1/8int(s^4+2s^2+1)/(s^3)ds`

`color(white)(I)=1/8ints+2s^-1+s^-3ds`

`color(white)(I)=1/8(1/2s^2+2ln(s)-1/2s^-2)+C`

`color(white)(I)=1/16(s^2+4ln(s)-1/(s^2))+C`

Substitute back `s=tan(u/2)` and `u=2x`

`I=1/16(tan^2(x)+4ln(tan(x))-1/(tan^2(x)))+C`

Answer 2

`I=-1/4[csc(2x)cot(2x)+ln|csc(2x)+cot(2x)|]+c`

Explanation:

Here,

`I=intcsc^3 2x dx`

`=int(csc2x)(csc^2(2x))dx`

`=int(sqrt(cot^2(2x)+1))csc^2(2x)dx`

Let,

`color(blue)(cot(2x)=u)=>-csc^2(2x)*2dx=du`

`=>csc^2(2x)dx=-1/2du`

`I=intsqrt(u^2+1)(-1/2)du`

We know that,

`color(red)(intsqrt(x^2+a^2)dx=x/2sqrt(x^2+a^2)+a^2/2ln|x+sqrt(x^2+a^2)|+c`

`I=-1/2intsqrt(u^2+1^2)du`

`=-1/2[u/2sqrt(u^2+1)+1/2ln|u+sqrt(u^2+1)|]+c`

substituting back `color(blue)(u=cot2x`

#=-1/2[cot(2x)/2sqrt(cot^2(2x)+1)+1/2ln|cot(2x)+sqrt(cot^2(2x)+1 )|]+c#

`=-1/2[cot(2x)/2csc(2x)+1/2ln|cot(2x)+csc(2x)|]+c`

`=-1/4[csc(2x)cot(2x)+ln|csc(2x)+cot(2x)|]+c`