Evaluate the definite integral sin^5(x) cos^8(x) dx from 0 to pi/2

do I need to rewrite the sin^5(x) in terms of cos? or this integration by parts?

please help.

∫sin^5(x)*cos^8(x) dx from 0 to pi/2

∫sin(x)*sin^4(x)*cos^8(x) dx

Now re-write sin^4(x) in terms of cos(x) so we can use a substitution to integrate.

∫sin(x)(1 - cos²(x))²*cos^8(x) dx

u = cos(x)

-du = sin(x) dx

-∫(1 - 2u² + u^4)(u^8) du eval. from 1 to 0

= [-u^9/9 + 2u^11/11 - u^13/13 eval. from 1 to 0]

= 1/9 - 2/11 + 1/13 = 8/1287

You need to re-write sin^5(x) to "save a sine" as follows:
∫ sin^5(x)cos^8(x) dx (from x=0 to π/2)
= ∫ sin^4(x)cos^8(x) [sin(x) dx] (from x=0 to π/2), by pulling out the extra sin(x)
= ∫ cos^8(x)[1 - cos^2(x)]^2 [sin(x) dx] (from x=0 to π/2), since sin^2(x) = 1 - cos^2(x).

At this point, substitute:
u = cos(x) ==> du = -sin(x) dx.

By changing the bounds:
Lower bound: x = 0 ==> u = cos(0) = 1
Upper bound: x = 1 ==> u = cos(π/2) = 0.
(Note that the lower bound is greater than the upper. This is fine!)

This yields:
∫ cos^8(x)[1 - cos^2(x)]^2 [sin(x) dx] (from x=0 to π/2)
= ∫ -u^8(1 - u^2)^2 du (from u=1 to 0), by applying substitutions
= ∫ u^8(1 - u^2)^2 du (from u=0 to 1)
(Note that - ∫ f(x) dx (from x=a to b) = ∫ f(x) dx (from x=b to a).)
= ∫ u^8(u^4 - 2u^2 + 1) du (from u=0 to 1)
= ∫ (u^12 - 2u^10 + u^8) du (from u=0 to 1)
= [(1/13)u^13 - (2/11)u^11 + (1/9)u^9] (evaluated from u=0 to 1)
= (1/13 - 2/11 + 1/9) - (0 - 0 + 0)
= 8/1287.
(Verified: http://www.wolframalpha.com/input/?i=%E2… )

I hope this helps!