How do you integrate?

Question

#intsin(cos^2 2x)*sin4x dx#

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Answer 1 · 2018-04-07T16:28:41

#I=1/2cos(1/2+1/2cos(4x))+C#

We want to integrate

#I=intsin(cos^2(2x))*sin(4x)dx#

Remember the identity

#color(blue)(cos^2(a)=1/2+1/2cos(2a)#

Thus

#I=intsin(1/2+1/2cos(4x))*sin(4x)dx#

Make a substitution #u=1/2+1/2cos(4x)=>du=-2sin(4x)dx#

#I=-1/2intsin(u)du#

#I=1/2cos(u)+C#

Substitute back #u=1/2+1/2cos(4x)#

#I=1/2cos(1/2+1/2cos(4x))+C#