This integral is pretty tricky. It's going to require the use of a few trigonometric identities and rules for integration. I'll include definitions or explanations of the rules used at the very end in the case that you would find this helpful.
I would begin by using half-angle identities:
#sin^2(theta)=1/2(1-cos(2theta))#
#cos^2(theta)=1/2(1+cos(2theta))#
Note that these are not the original half-angle identities, but the originals have been manipulated to produce these. I will show that at the end if you are interested.
#intsin^4(x)*cos^2(x)dx#
#=>int(1/2(1-cos(2x))*1/2(1-cos(2x))*1/2(1+cos(2x)))dx#
Factor out #1/8# and multiply out:
#=>1/8int(cos^2(2x)-2cos(2x)+1)(1+cos(2x))dx#
#=>1/8int(color(green)(cos^2(2x))color(red)(-2cos(2x))+1+cos^3(2x)color(green)(-2cos^2(2x))+color(red)(cos(2x))dx#
Combine like terms:
#=>1/8int(-cos(2x)+1+cos^3(2x)-cos^2(2x))dx#
#=> 1/8int(cos^3(2x)-cos^2(2x)-cos(2x)+1)dx#
We can use the sum rule to rewrite as
#=>1/8int(cos^3(2x))dx-1/8int(cos^2(2x))dx-1/8int(cos(2x))dx+1/8int(1)dx#
The last integral, #1/8int(1)dx# can be easily taken care of without any further manipulation.
#1/8int1dx=>1/8x#
Note: I will leave out adding the constant, #C# until the end.
#-1/8intcos(2x)dx# can also be taken care of fairly easily, this time with a simple #u# substitution.
#-1/8intcos(2x)dx#
#u=2x => du=2dx => 1/2du=dx#
#=>-1/8*1/2intcos(u)du#
#=>-1/16sin(u)#
#=>-1/16sin(2x)#
This leaves us with
#1/8int(cos^3(2x))dx-1/8int(cos^2(2x))dx#
We can use the half-angle identity for cosine again to work out the second integral after doing a #u# substitution. It isn't strictly necessary to do the substitution, but I'll show it to avoid any confusion.
#-1/8intcos^2(2x)dx#
#u=2x => du=2dx => 1/2du=dx#
#=>-1/8int1/2cos^2(u)du#
Now apply the half-angle identity:
#=>-1/16int1/2(1+cos(2u))dx#
#=>-1/32int(1+cos(2u))du#
Apply sum rule:
#=>-1/32int1du-1/32intcos(2u)du#
Again, you may show a substitution here, but it is not strictly necessary. You may avoid the sum rule and substitution and integrate in one step if you are comfortable doing so. I'll use #z# as a variable since we've already used #u#.
#z=2u => dz=2du =>1/2dz=du#
#=>-1/32int1du-1/32int(1/2*cos(z))du#
#=>-1/32u-1/64sin(z)#
Substitute back in for #z#
#=>-1/32u-1/64sin(2u)#
Substitute back in for #u#
#=>-1/32(2x)-1/64sin(2(2x))#
Simplify
#=>-1/16x-1/64sin(4x)#
So far we have:
#color(red)(1/8x)-1/16sin(2x)color(red)(-1/16x)-1/64sin(4x)+1/8int(cos^3(2x))dx#
We can simplify:
#=>x/16-1/16sin(2x)-1/64sin(4x)+1/8int(cos^3(2x))dx#
And now we're left to solve
#1/8int(cos^3(2x))dx#
Rewrite and apply Pythagorean identity:
#=>1/8intcos(2x)*cos^2(2x)dx#
#=>1/8intcos(2x)*(1-sin^2(2x))dx#
Apply #u# substitution:
#u=sin(2x) => du=2cos(2x)dx => 1/2du=cos(2x)#
#=>1/8int1/2(1-u^2)du#
#=>1/16int(1-u^2)du#
Integrate
#=>1/16(u-1/3u^3)#
Substitute back in for #u#:
#=>1/16(sin(2x)-1/3sin^3(2x))#
Putting it all together, we have:
#intsin^4(x)*cos^2(x)dx=x/16-1/16sin(2x)-1/64sin(4x)+1/16(sin(2x)-1/3sin^3(2x))+C#
We can simplify:
#=>x/16color(red)(-1/16sin(2x))-1/64sin(4x)+color(red)(1/16sin(2x))-1/48sin^3(2x)+C#
#=>x/16-1/64sin(4x)-1/48sin^3(2x)+C#
Our final answer is then
#x/16-sin(4x)/64-sin^3(2x)/48+C#
Sum rule:
#intf(x)+g(x)dx=intf(x)dx+intg(x)dx#
Pythagorean identity:
#sin^2(theta)+cos^2(theta)=1#
Half Angle identities:
#sin(x/2)=+-sqrt((1-cos(x))/2)#
#cos(x/2)=+-sqrt((1+cos(x))/2)#
If we square each side and double all of the angle measures, we get:
#sin^2(theta)=1/2(1-cos(2theta))#
#cos^2(theta)=1/2(1+cos(2theta))#
