How to evaluate $\int \cos^2x \ dx$

Question

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.

Answer 1

You can make use of the double angle formula for cosine:

$$\begin{eqnarray}\cos(2x) & = & 2 \cos^2 (x) - 1 \\ \cos^2 (x) & = & \frac{1}{2}\left(\cos (2x) + 1\right)\\ \int \cos^2 x\ dx & = & \frac{1}{2} \int \left(\cos(2x) + 1 \right) dx \\ & = & \frac{1}{2}\left[\frac{1}{2}\sin(2x) + x + C \right] \\ & = & \frac{1}{4}\sin(2x) + \frac{1}{2}x + C \end{eqnarray}$$

Answer 2

Hint : Use the identity $\cos 2x=2\cos^2 x-1$

i.e. $$\cos^2 x=\frac{\cos 2x +1}{2}$$

Answer 3

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 $$\int \cos^2 x \, dx = \frac{1}{2} (\cos x \cdot \sin x + x + C)$$ $$=\frac{1}{4}\sin 2x + \frac{1}{2} x + C'$$

Answer 4

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.

Answer 5

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 $$2I=x+\frac 12 \sin(2x)$$ $$I=\frac x2+\frac 14 \sin(2x)$$

Answer 6

The substitution $$\cos x=\frac{e^{ix}+e^{-ix}}{2}$$ make it easy to evaluate. $$\int \cos^2 x\,dx=\frac14\int(e^{2ix}+e^{-2ix}+2)\,dx=\frac14\Big(\sin(2x)+2x\Big)$$