How to evaluate $\int \cos^2x \ dx$ I am new to integration and I want to evaluate
$ \int \cos^2x \, dx $
What I done
Using simple chain rule,
$ \int(\cos x)^2 \\
= \int t^2 \,dt  \ (t = \cos x) \\
= \frac{t^3}{3} \ + C $
By substitution of $t = \cos(x)$ I got with answer.
 A: Consider $$I=\int \cos^2(x)\,dx \qquad J=\int \sin^2(x)\,dx $$ So $$I+J =\int dx=x\tag 1$$ $$I-J =\int \left(\cos^2(x)-\sin^2(x) \right) dx=\int \cos(2x) \,dx=\frac 12 \sin(2x)\tag 2$$ Add both to get $$2I=x+\frac 12 \sin(2x)$$ $$I=\frac x2+\frac 14 \sin(2x)$$
A: You can make use of the double angle formula for cosine:
$$\begin{eqnarray}\cos(2x) & = & 2 \cos^2 (x) - 1 \\
\cos^2 (x) & = & \frac{1}{2}\left(\cos (2x) + 1\right)\\
\int \cos^2 x\ dx  & = & \frac{1}{2} \int \left(\cos(2x) + 1 \right) dx \\
& = & \frac{1}{2}\left[\frac{1}{2}\sin(2x) + x + C \right] \\
& = & \frac{1}{4}\sin(2x) + \frac{1}{2}x + C \end{eqnarray}$$
A: Hint : Use the identity $\cos 2x=2\cos^2 x-1$
i.e. $$\cos^2 x=\frac{\cos 2x +1}{2}$$
A: Another way is to use integration by parts:
$$ \int \cos^2x \, dx $$
$$=\int \cos x \cdot \cos x \, dx $$
$$=\int \cos x \,d(\sin x)$$
$$=\cos x \cdot \sin x -\int \sin x \, d(\cos x) $$
$$=\cos x \cdot \sin x +\int \sin^2 x \, dx $$
So
$$\int \cos^2x \, dx =\cos x \cdot \sin x +\int (1-  \cos^2 x) \, dx $$
$$=\cos x \cdot \sin x + x + C-\int \cos^2 x \, dx $$
Thus 
$$\int \cos^2 x \, dx = \frac{1}{2} (\cos x \cdot \sin x + x + C)$$
$$=\frac{1}{4}\sin 2x + \frac{1}{2} x + C'$$
A: This is very bad way.
Take $\sin x=t$, then $\cos x dx=dt$. Substitute and you get $$\int \sqrt{1-t^2}dt$$
If you know Integration by parts, you can do next steps easily.
A: The substitution $$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$ make it easy to evaluate.
$$\int \cos^2 x\,dx=\frac14\int(e^{2ix}+e^{-2ix}+2)\,dx=\frac14\Big(\sin(2x)+2x\Big)$$
