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What is the integral of #int cos^3 (x)dx# from 0 to pi/2?

1 Answer

Feb 8, 2017

#int_0^(pi/2) cos^3xdx = 2/3#

Explanation:

Use the idendity #cos^2x+sin^2x = 1#:

#int_0^(pi/2) cos^3xdx = int_0^(pi/2) cosx cos^2xdx = int_0^(pi/2) cosx (1-sin^2x) xdx#

substitute:

#t=sinx#
#dt = cosx dx#
#x in (0,pi/2) => t in (0,1)#

so:

#int_0^(pi/2) cos^3xdx = int_0^1 (1-t^2)dt = [t-t^3/3]_0^1 = 2/3#

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