Three Integrals

[ChaoticLlama]

I'm doing some work out of a book a picked up this week to give my brain a refresher on basic integration techniques, and I have forgotten how to do some of the following.

∫(sinx)/(1 + cos²x) dx

∫x³√(1 - x²) dx

∫ (x^7)/(1 - x^4)² dx

Thanks for any hints/solutions!

 

[tkhunny]

Substitute u = 1 + [cos(x)]², du = -2sin(x) dx

Substitute u = 1 - x², x² = 1-u, du = -2x dx

Substitute u = 1 - x⁴, x⁴ = 1-u, du = -4x³ dx

Hmmm...This must be the "substitution" section.

 

[stapel]

1) Let u = cos(x), so -sin(x)dx = du. Then check your arc-function derivative formulas.

2) Let u = 1 - x², so du = -2xdx and x² = 1 - u. Note that (u)(sqrt(u)) = u^3/2. Use the power rule for integration.

3) Let u = 1 - x⁴, so du = -4x³dx, and x⁷ = (x⁴)(x³). Split the integral into two pieces; you'll need logs for one of the pieces.

Eliz.

Edit: Ah, ya beat me by two minutes! :wink: