#int\ sin^4(x)\ dx=3/8x-1/4sin(2x)+1/32sin(4x)+C#

This integral is mostly about clever rewriting of your functions. As a rule of thumb, if the power is even, we use the double angle formula. The double angle formula says:
#sin^2(theta)=1/2(1-cos(2theta))#

If we split up our integral like this,
#int\ sin^2(x)*sin^2(x)\ dx#

We can use the double angle formula twice:
#int\ 1/2(1-cos(2x))*1/2(1-cos(2x))\ dx#

Both parts are the same, so we can just put it as a square:
#int\ (1/2(1-cos(2x)))^2\ dx#

Expanding, we get:
#int\ 1/4(1-2cos(2x)+cos^2(2x))\ dx#

We can then use the other double angle formula
#cos^2(theta)=1/2(1+cos(2theta))#
to rewrite the last term as follows:
#1/4int\ 1-2cos(2x)+1/2(1+cos(4x))\ dx=#

#=1/4(int\ 1\ dx-int\ 2cos(2x)\ dx+1/2int\ 1+cos(4x)\ dx)=#

#=1/4(x-int\ 2cos(2x)\ dx+1/2(x+int\ cos(4x)\ dx))#

I will call the left integral in the parenthesis Integral 1, and the right on Integral 2.

Integral 1
#int\ 2cos(2x)\ dx#

Looking at the integral, we have the derivative of the inside, #2# outside of the function, and this should immediately ring a bell that you should use u-substitution.

If we let #u=2x#, the derivative becomes #2#, so we divide through by #2# to integrate with respect to #u#:
#int\ (cancel(2)cos(u))/cancel(2)\ du#

#int\ cos(u)\ du=sin(u)=sin(2x)#

Integral 2
#int\ cos(4x)\ dx#

It's not as obvious here, but we can also use u-substitution here. We can let #u=4x#, and the derivative will be #4#:
#1/4int\ cos(u)\ dx=1/4sin(u)=1/4sin(4x)#

Completing the original integral
Now that we know Integral 1 and Integral 2, we can plug them back into our original expression to get the final answer:
#1/4(x-sin(2x)+1/2(x+1/4sin(4x)))+C=#

#=1/4(x-sin(2x)+1/2x+1/8sin(4x))+C=#

#=1/4x-1/4sin(2x)+1/8x+1/32sin(4x)+C=#

#=3/8x-1/4sin(2x)+1/32sin(4x)+C#