Explanation:
The definite integral of sin^5(x)dx from 0 to pi/2 can be evaluated using the method of substitution.
Let u = sin(x), then du = cos(x)dx
The integral becomes:
∫sin^5(x)dx = ∫u^5du from 0 to sin(π/2)
= (u^6)/6 evaluated at sin(π/2) and 0
= (sin^6(π/2))/6 - 0
= (1^6)/6
= 1/6
So, the definite integral of sin^5(x)dx from 0 to pi/2 is equal to 1/6.