Find $\int \cos^4(x)dx$ We have: $\int \cos^n x\ dx = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n}\int \cos^{n-2} x\ dx.$
Find $\int \cos^4x\ dx$ by using the formula twice
What I have so far is:
$\int \cos^4 x\ dx = \frac{1}{4} \cos^{3} x \sin x + \frac{3}{4}\int \cos^{2} x\
 dx$
Now we use the formula for $\int cos^{2} x\ dx$:
$\int \cos^2 x\ dx = \frac{1}{2} \cos x \sin x + \frac{1}{2}\int \cos^{0} x\
 dx$
$\int \cos^2 x\ dx = \frac{1}{2} \cos x \sin x + \frac{1}{2}\int 1 \
 dx$
$\int \cos^2 x\ dx = \frac{1}{2} \cos x \sin x + \frac{1}{2}[ x ]$
$\int \cos^2 x\ dx = \frac{1}{2} \cos x \sin x + \frac{x}{2}$
Now plug this in to the  $\int \cos^4 x\ dx$ equation above
$\int \cos^4 x\ dx = \frac{1}{4} \cos^{3} x \sin x + \frac{3}{4}[\frac{1}{2} \cos x \sin x + \frac{x}{2}]$
$\int \cos^4 x\ dx = \frac{1}{4} \cos^{3} x \sin x + \frac{3}{8}[\cos x \sin x + {x}]$
This is where I get stuck. I'm aware I could have used an identity for $\cos^2x$ but the question needs me to use the formula twice.

You are right.
Look at the alternate forms in WolframAlpha

