# 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.

• $$dt \neq dx$$ – The Chaz 2.0 Jul 7 '17 at 4:33
• @The Chaz 2.0 So what should be the correct thing? – Ravi Prakash Jul 7 '17 at 4:37
• check out this example – The Chaz 2.0 Jul 7 '17 at 4:42
• @the chaz 2.0 I'm integrating with respect to $t$ , so why I can't write there $dt$ ? – Ravi Prakash Jul 7 '17 at 4:49
• – projectilemotion Jul 7 '17 at 8:04

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

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

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

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

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

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.

• I don't know integration by parts – Ravi Prakash Jul 7 '17 at 4:47
• @RaviPrakash you started integration recently isn't!! You will learn Integration by parts very soon, you can solve it very easily in this process as soon as you know the method. – MAN-MADE Jul 7 '17 at 4:51

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

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