Why can't I do $u$ substitution on $\int \cos^4x dx$? I tried to rewrite the integral as $$\int (\cos^2x)(\cos^2x)dx$$ which in turn I used the trig identity of $\cos^2x = 1 - \sin^2x$ and therefore changed $$\int (\cos^2x)(\cos^2x)dx = \int (1 - \sin^2x)(1 - \sin^2x)dx = \int ( \sin^4x - \sin^2x + 1) dx$$
I then tried to do u substitution where $u=\sin x$ and $du=\cos x dx$
But my answer was completely wrong...I ended up getting $$\frac{\sin^5x}{5}+\sin x-\frac{2\cos^3x}{3}+K, \quad K\in \mathbb R$$
Am I doing $u$ substitution wrong or is there a better way to solve this integral?
 A: If I were grading your work, I'd deduct a point for leaving the $dx$ and $du$ out of the integral.  When you replace $dx$ by $du/\cos x$, you'll have a much different integrand.   I don't think there's a $u$-substitution way to work this problem.
The usual way is to use the identity 
$$\cos^2x = \frac{1+\cos 2x}{2}.$$
Replace each of your $cos^2 x$  and multiply.  Then you'll have a $\cos^2 2x$ which you can replace using the same identity.
A: Your substitution is not correct. Here are some steps for solving given integral:
$$\cos^4x = (\cos^2x)^2, \quad \cos^2x = \frac{1+\cos2x}{2}.$$
Thus, we can write:
$$\cos^4x = \frac{1 + 2\cos2x + \cos^22x}{4}=\frac{1}{4}+\frac{1}{2}\cos2x+\frac{1}{4}\cos^22x.$$
Now, we will apply once again a formula for square of cosines:
$$\cos^4x = \frac{1}{4}+\frac{1}{2}\cos2x+\frac{1}{4}(\frac{1+\cos4x}{2})=\frac{1}{4}+\frac{1}{2}\cos2x+\frac{1}{8}+\frac{1}{8}\cos4x.$$
The solution for given integral is now easy to obrain...
A: $$\int_{ }^{ }\cos^{4}\left(x\right)dx=\frac{1}{4}\int_{ }^{ }\left(1+\cos\left(2x\right)\right)^{2}dx$$$$=\frac{1}{4}\left[\int_{ }^{ }dx+2\int_{ }^{ }\cos\left(2x\right)dx+\int_{ }^{ }\cos^{2}\left(2x\right)dx\right]$$$$=\frac{1}{4}\left[\int_{ }^{ }dx+2\int_{ }^{ }\cos\left(2x\right)dx+\frac{1}{2}\int_{ }^{ }1+\cos\left(4x\right)dx\right]$$$$=\frac{1}{4}\left[\int_{ }^{ }dx+2\int_{ }^{ }\cos\left(2x\right)dx+\frac{1}{2}\left(\int_{ }^{ }dx+\int_{ }^{ }\cos\left(4x\right)dx\right)\right]$$$$=\frac{1}{4}\left[x+\sin\left(2x\right)+\frac{1}{2}\left(x+\frac{1}{4}\sin\left(4x\right)\right)\right]+C$$
