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

1 Answer
Øko
Apr 15, 2018

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

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