Trig Integrals?

Question

Hello -

I have seem to forgotten how to do some fairly simple trig integrals.

Does anyone know how to do to the following integral?

Integral (cos^2(t)sin(t))dt

I know the answer is - (cos^3(t))/3, but I am no sure how to get it.

Please help....thanks.

Answer 1

I = ∫ (cos ² t) (sin t) dt
let u = cos t
du = - sin t dt
- du = sin t dt
I = - ∫ u ² du
I = - u ³ / 3 + C
I = - cos ³ t / 3 + C

Answer 2

∫cos^2(t) sin(t) dt

let cos(t) = u

-sin(t)dt = du

sin(t) dt = -du

Now the integral becomes

∫u^2 (-du) = -∫u^2 du = -u^3/3 + c = (1/3)cos^3(t)+c

Answer 3

∫ (cos^2(t)sin(t))dt
= ∫ (-cos^2(t))dcos(t)
= - (cos^3(t))/3 + c
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Ideas: Mental substitution sin(t) dt = -dcos(t)