Asked by Anonymous
Find an antiderivative
∫t^2sin^3(t^3) cos^2(t^3) dt.
∫t^2sin^3(t^3) cos^2(t^3) dt.
Answers
Answered by
oobleck
∫t^2sin^3(t^3) cos^2(t^3) dt
Let u = t^3, so du = 3t^2 dt, and that gives you
1/3 ∫sin^3(u) cos^2(u) du
= 1/3 ∫ sinu (1-cos^2u)(cos^2u) du
= -1/3 ∫ cos^2u - cos^4u (-sinu du)
Now let v = cosu, so dv = -sinu du, and that gives you
-1/3 ∫ v^2 - v^4 dv
= -1/3 (1/3 v^3 - 1/5 v^5) + C
= 1/15 cos^5(t^3) - 1/9 cos^3(t^3) + C
Let u = t^3, so du = 3t^2 dt, and that gives you
1/3 ∫sin^3(u) cos^2(u) du
= 1/3 ∫ sinu (1-cos^2u)(cos^2u) du
= -1/3 ∫ cos^2u - cos^4u (-sinu du)
Now let v = cosu, so dv = -sinu du, and that gives you
-1/3 ∫ v^2 - v^4 dv
= -1/3 (1/3 v^3 - 1/5 v^5) + C
= 1/15 cos^5(t^3) - 1/9 cos^3(t^3) + C