How do you evaluate the definite integral #int sin^3xdx# from #[pi/3, pi/2]#?

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Answer 1 · 2017-08-03T11:04:30

# 11/24.#

Let, #I=int_(pi/3)^(pi/2)sin^3xdx.#

We know that, #sin3x=3sinx-4sin^3x.#

#rArr sin^3x=(3sinx-sin3x)/4.#

#:. I=int_(pi/3)^(pi/2) (3sinx-sin3x)/4dx,#

#=1/4[-3cosx-(-cos(3x)/3)]_(pi/3)^(pi/2),#

#=1/12[cos3x-9cosx]_(pi/3)^(pi/2),#

#=1/12[{cos(3pi/2)-9cos(pi/2)}-{cos(3pi/3)-(9cos(pi/3)}},#

#=1/12[0-{-1-9*1/2}],#

# rArr I=11/24.#

Answer 2 · 2017-08-03T11:36:07

#int_(1/3pi)^(1/2pi)sin^3xdx=11/24#

Let #f^((-1))(x)=intsin_(1/3pi)^(1/2pi)dx#

Use the reduction formula:

#intsin^nxdx=-1/nsin^(n-1)xcosx +(n-1)/n intsin^(n-2)dx#

Then #f^((-1))(x)=[-1/3sin^2xcosx]_(1/3pi)^(1/2pi)+2/3int_(1/3pi)^(1/2pi)sinxdx#

I shall leave the subsequent evaluation of the integral up to you as an exercise.