Recent content by loserspearl

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∫(x2 + 7x) cosx dx If I make v = (x2 + 7x) and du = cosx dx I get ((x2 + 7x) sinx)/2 If I make v = cosx and du = (x2 + 7x) dx I get ((x3/3 + 7x2/2) cosx)/2 using the form X=Y-X to X=Y/2 Neither are correct, what did I do wrong?

Huh? What did you do after -1/32 ∫(1-u2)2 du? Not sure what you mean about my error, pi/2 is the limit why would it be 2?

So v=sinθ and du=cos dθ makes dv=cosθ (or is it cosθ dθ?) and u=sinθ sin5θ * sin5θ - ∫sin5θ * cos5θ dθ ?? because I'm back at the same problem I even tried the X=Y-X form and changed in 2X=Y/2 and still didn't get the right answer

from 0 to π/2 ∫sin5θ cos5θ dθ I have been trying to solve the above for quite some time now yet can't see what I am doing wrong. I break it down using double angle formulas into: ∫ 1/25 sin5(2θ) dθ 1/32 ∫sin4(2θ) * sin(2θ) dθ 1/32 ∫(1-cos2(2θ))2 * sin(2θ) dθ With this I can make u = cos(2θ)...