Applying the reduction formula with `n = 4` gives:

`int sin^4x\ dx` `=-1/4cos x\ sin^3x+3/4I_2`

We now need to find `I_2=intsin^2x\ dx`, which corresponds to `n=2`.

Now, from our table of integrals,

`intsin^2xdx` `=x/2-1/2sin x\ cos x+K`

So putting it all together gives:

`intsin^4x\ dx`

`=-1/4cos x\ sin^3x` `+3/4[x/2-1/2sin x\ cos x+K]`

`=-1/4cos x\ sin^3x` `+3/8x` ` -3/8 cos x\ sin x+K^'`

`=3/8 x - 1/4 sin(2 x)` `+ 1/32 sin(4 x)+K^'` (afer some manipulation)

NOTE: We used `K` and `K^'` since the value of those constants is actually different.