# Calculate int_0^(2pi) cos^2x .dx?

## A) Zero B) 2 C) 1/2 D) $\pi$ The answer is D for your reference

Sep 5, 2017

Yes, the correct answer is $D$, or $\pi$.

#### Explanation:

Use the power reduction formula

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

So we have:

$I = {\int}_{0}^{2 \pi} \frac{1 + \cos 2 x}{2}$

$I = \frac{1}{2} {\int}_{0}^{2 \pi} 1 + \cos \left(2 x\right) \mathrm{dx}$

$I = \frac{1}{2} {\int}_{0}^{2 \pi} 1 \mathrm{dx} + \frac{1}{2} {\int}_{0}^{2 \pi} \cos \left(2 x\right) \mathrm{dx}$

If we let $u = 2 x$ in the second integral, then $\frac{1}{2} \mathrm{du} = \mathrm{dx}$. The integral of $\cos u$ is $\sin u$ and reverse your substitution to get $\sin \left(2 x\right)$.

$I = \frac{1}{2} {\left[x\right]}_{0}^{2 \pi} + \frac{1}{2} \left(\frac{1}{2}\right) {\left[\sin \left(2 x\right)\right]}_{0}^{2 \pi}$

The integral equals

$I = \frac{1}{2} \left(2 \pi\right) + \frac{1}{4} \left(\sin \left(4 \pi\right) - \sin \left(0\right)\right)$

$I = \pi - \frac{1}{4} \left(0 - 0\right)$

$I = \pi$

Answer $D$, as required.

Hopefully this helps!

Sep 6, 2017

${\int}_{0}^{2 \pi} {\cos}^{2} \left(x\right) \mathrm{dx}$

We can rewrite this:

${\int}_{0}^{2 \pi} {\cos}^{2} \left(x\right) \mathrm{dx} = {\int}_{0}^{2 \pi} \left({\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right) + {\sin}^{2} \left(x\right)\right) \mathrm{dx}$

Note that $\cos \left(2 x\right) = {\cos}^{2} \left(x\right) - {\sin}^{2} \left(x\right)$:

${\int}_{0}^{2 \pi} {\cos}^{2} \left(x\right) \mathrm{dx} = {\int}_{0}^{2 \pi} \cos \left(2 x\right) \mathrm{dx} + {\int}_{0}^{2 \pi} {\sin}^{2} \left(x\right) \mathrm{dx}$

Rewrite ${\sin}^{2} \left(x\right)$ using ${\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right) = 1$:

${\int}_{0}^{2 \pi} {\cos}^{2} \left(x\right) \mathrm{dx} = {\int}_{0}^{2 \pi} \cos \left(2 x\right) \mathrm{dx} + {\int}_{0}^{2 \pi} \left(1 - {\cos}^{2} \left(x\right)\right) \mathrm{dx}$

${\int}_{0}^{2 \pi} {\cos}^{2} \left(x\right) \mathrm{dx} = {\int}_{0}^{2 \pi} \cos \left(2 x\right) \mathrm{dx} + {\int}_{0}^{2 \pi} \mathrm{dx} - {\int}_{0}^{2 \pi} {\cos}^{2} \left(x\right) \mathrm{dx}$

Add ${\int}_{0}^{2 \pi} {\cos}^{2} \left(x\right) \mathrm{dx}$ to both sides of the equation:

$2 {\int}_{0}^{2 \pi} {\cos}^{2} \left(x\right) \mathrm{dx} = {\int}_{0}^{2 \pi} \cos \left(2 x\right) \mathrm{dx} + {\int}_{0}^{2 \pi} \mathrm{dx}$

Evaluate these integrals:

$2 {\int}_{0}^{2 \pi} {\cos}^{2} \left(x\right) \mathrm{dx} = | \frac{1}{2} \sin \left(2 x\right) {|}_{0}^{2 \pi} + x {|}_{0}^{2 \pi}$

$2 {\int}_{0}^{2 \pi} {\cos}^{2} \left(x\right) \mathrm{dx} = \frac{1}{2} \left(\sin \left(4 \pi\right) - \sin \left(0\right)\right) + \left(2 \pi - 0\right)$

$2 {\int}_{0}^{2 \pi} {\cos}^{2} \left(x\right) \mathrm{dx} = 2 \pi$

Dividing by $2$:

${\int}_{0}^{2 \pi} {\cos}^{2} \left(x\right) \mathrm{dx} = \pi$