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))
