$\int_0^\pi \cos^2 x$ - Where did I go wrong? So when looking at the question:
$$\int_{0}^{\pi} \cos^2 x \ \text{d}x$$
I would just subtract $\cos^2(0)$ from $\cos^2(\pi)$, but doing so would get me 1 - 1 = 0. When the answer is $\pi/2$. Where did I go wrong? What am I missing? Thanks so much for all your help! :-)
 A: From the addition identity:
$$\cos (a+b)=\cos a\cdot \cos b-\sin a\cdot \sin b,$$
we get (setting $a=b$)
$$\cos (2a)=\cos ^{2}a-\sin ^{2}a.$$
Applying the Pythagorean trigonometric identity $\cos^2a+\sin^2a=1$, in the form $$\sin^2a=1-\cos^2a,$$
yields 
$$\cos (2a)=\cos ^{2}a-\sin ^{2}a=\cos ^{2}a-1+\cos^2a=2\cos ^{2}a-1,$$
or, equivalently
$$\cos ^{2}a=\dfrac{1+\cos (2a)}{2}.$$
Setting $x=a$ results in
$$\cos ^{2}(x)=\dfrac{1+\cos (2x)}{2}=\dfrac{1}{2}+\dfrac{\cos (2x)}{2}.$$
Then 
$$\int_{0}^{\pi} \cos^2 x \ \text{d}x=\int_{0}^{\pi}\dfrac{1}{2}+\dfrac{\cos (2x)}{2}  \ \text{d}x=\dfrac{1}{2}\pi+\dfrac{1}{2}\int_{0}^{\pi}\cos (2x)\ \text{d}x=\dfrac{1}{2}\pi+\dfrac{1}{4}\int_{0}^{2\pi }\cos t\;\mathrm{d}t.$$
I leave to you the evaluation of $\displaystyle\int_{0}^{2\pi}\cos t\ \text{d}t$. Remember that you have to find the antiderivative of $\cos t$, or just observe that the period of $\cos t$ is equal to $2\pi$.
A: We have that
$$I = \int_{0}^{\pi} \cos^2 x \ dx= 2 \int_{0}^{\pi/2} \cos^2 x \ dx = 2 \int_{0}^{\pi/2} \cos^2 (\pi/2 - x) \ dx =  2 \int_{0}^{\pi/2} \sin^2 x \ dx$$
and thus
$$I = \int_{0}^{\pi/2} (\cos^2 x +\sin^2 x)\ dx = \pi/2$$
A: You need to integrate the integrand $\cos^2(x)$ first. The identity $\displaystyle\cos^2(x)=\frac{1+\cos(2x)}{2}$ is of use here.
