what's integral of sin^2x cosx dx from o to pi/2?

Question

sin^3 x /3 ?

Answer 1

Nopes the ans has to be a real number as there are limits to the integral.

If you substitute sinx = a and then integrate, the answer that is got is 1/3.

Method :
put sin x =a

therefore, da = cosx dx

therefore, the integral becomes

integral of a^2 da from 0 to 1.

I guess its simple from there onwards

Answer 2

Let u = sin x
Then du = cos x dx
So we have integral u^2du = u^3/3 = 1/3sin^3 x
from 0 to pi/2 = 1/3

Answer 3

(1/3)sin^3 x...now evaluate at pi/2 and at 0 and subtract to get 1/3.

Answer 4

∫sin^2(x) cos(x) dx

let sin x = y

cos(x)dx = dy

∫y^2 dy

=(y^3/3)

=>(1/3[sin^3(x)]

between 0 and pi/2

(1/3)[sin^3(pi/2) - sin^3(0)]

=>(1/3)[(1)^3 - 0]

=> 1/3

Answer 5

∫ u^n du = u^(n+1)/(n+1) + C

π/2
∫ sin^2x cosx dx
0

1/3*sin^3 (π/2) - 1/3sin^3(0) = 1/3

Answer 6

zero