Question

How do you find the indefinite integral of ∫ dx / sin^3 x cos^5 x ?

Answer 1

`I=1/4tan^4(x)+3/2tan^2(x)+3ln(tan(x))-1/2cot^2(x)+C`

Explanation:

We want to solve

`I=int1/(sin^3(x)cos^5(x))dx`

Multiply the DEN and NUM by `sec^8(x)`,
an integral consisting of tangens and secant have good opputunities substitution

`I=intsec^8(x)/(tan^3(x))dx`

Make a substitution `color(red)(u=tan(x)=>du=sec^2(x)dx`

`I=intsec^6(x)/(u^3)du`

Use the identity `color(blue)(sec^2(x)=tan^2(x)+1`

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

`color(white)(I)=int(u^6+3u^4+3u^2+1)/(u^3)du`

`color(white)(I)=intu^3+3u+3/u+1/u^3du`

`color(white)(I)=1/4u^4+3/2u^2+3ln(u)-1/(2u^2)+C`

Substitute back `u=tan(x)`

`I=1/4tan^4(x)+3/2tan^2(x)+3ln(tan(x))-1/2cot^2(x)+C`