What is the integral of #int sin^3xcos^2x dx# from #0# to #pi/2#?

Question

8024 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-06T20:17:10

#2/15#

Rewrite the integrand as #int_0^(pi/2)sin^2xcos^2xsinxdx#

Recalling the identity #sin^2x=1-cos^2x,# we get

#int_0^(pi/2)(1-cos^2x)cos^2xsinxdx#

We can now solve this with a simple substitution.

#u=cosx#
#du=-sinxdx#

Calculate the new bounds:

Upper: #u=cos(pi/2)=0#
Lower: #u=cos0=1#

#-int_1^0u^2(1-u^2)du=-int_1^0(u^2-u^4)du#

#=-(1/3u^3-1/5u^5)|_1^0#
#-(-1/3+1/5)=2/15#