How do you integrate #int sin^2(x/5)*cos^3(x/5)#?

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Answer 1 · 2016-05-17T15:40:29

#5/3sin^3(x/5)-sin^5(x/5)+C#

We have

#intsin^2(x/5)cos^3(x/5)dx#

#intsin^2(x/5)cos^2(x/5)cos(x/5)dx#

Use #cos^2(x) = 1-sin^2(x)# to re write the expression as:

#intsin^2(x/5)(1-sin^2(x/5))cos(x/5)dx#

#=int(sin^2(x/5)-sin^4(x/5))cos(x/5)dx#

Now apply the substitution:
#u = sin(x/5)#
#->du = 1/5cos(x)dx#

Which will give us the integral:

#5intu^2-u^4du#

#=5(u^3/3-u^5/5)+C#

Now reverse the substitution to get:

#5/3sin^3(x/5)-sin^5(x/5)+C#