Question

What is the integral of `cos^4(x/2)`?

Answer 1

`1/16sin(2x)+1/2sin(x)+3/8x+C`

Explanation:

`I=intcos^4(x/2)`

We will use the cosine double-angle formula to rewrite this. The identity tells us that: `cos(2x)=2cos^2(x)-1`, which can me modified to say that `cos(x)=2cos^2(x/2)-1`. Solving for `cos^2(x/2)` yields `cos^2(x/2)=1/2(cos(x)+1)`. Squaring both sides yields `cos^4(x/2)=1/4(cos(x)+1)^2=1/4(cos^2(x)+2cos(x)+1)`.

`I=1/4int(cos^2(x)+2cos(x)+1)dx`

`I=1/4intcos^2(x)dx+1/2intcos(x)dx+1/4intdx`

The final two can be integrated easily:

`I=1/4intcos^2(x)dx+1/2sin(x)+1/4x`

The first integrand can be rewritten using the same formula as before: since `cos(2x)=2cos^2(x)-1`, we see that `cos^2(x)=1/2(cos(2x)+1)`. Thus:

`I=1/4int1/2(cos(2x)+1)dx+1/2sin(x)+1/4x`

`I=1/8intcos(2x)+1/8intdx+1/2sin(x)+1/4x`

`I=1/16sin(2x)+1/8x+1/2sin(x)+1/4x+C`

`I=1/16sin(2x)+1/2sin(x)+3/8x+C`