Definite integral $\int_{-\pi/2}^{\pi/2}\cos^{2}\left(\theta\right)\,{\rm d}{\theta} $ Please help me to evaluate definite integral
$$\int_{^{-\pi}/_2}^{^\pi/_2}\cos^{2}\left(\theta\right)\,{\rm d}{\theta}$$
Also there was a hint:
Use the double angle formula $\cos\left(2x\right)=2\cos^{2}\left(x\right) - 1$.
I know how to do an integral, but I don't know how to evaluate it with this hint. Please help me.
 A: Note that $$\int_{- \pi /2}^{\pi / 2} \cos^2 \theta d \theta= \int_{- \pi /2}^{\pi / 2} \sin^2 \theta d \theta \ $$ and that $$\pi = \int_{- \pi /2}^{\pi / 2} 1 d \theta =
\int_{- \pi /2}^{\pi / 2} (\cos^2 \theta +\sin^2 \theta) d \theta = 2 \int_{- \pi /2}^{\pi / 2} \cos^2 \theta d \theta$$
hence
$$\int_{- \pi /2}^{\pi / 2} \cos^2 \theta d \theta= \frac{\pi}{2}$$
A: From the hint, $\cos^2 x= \frac{1}{2}(\cos 2x+1)$ so our integral becomes $$\int_{-\pi/2}^{\pi/2} \frac{1}{2}(\cos 2x+1)dx= \left. \left (\frac{1}{4}\sin 2x+\frac{1}{2}x \right ) \right |_{-\pi/2}^{\pi/2}=\pi/2$$
A: $$
\int_{-\pi/2}^{\pi/2} \cos^2\theta\,d\theta + \int_{-\pi/2}^{\pi/2} \sin^2\theta\,d\theta = \int_{-\pi/2}^{\pi/2} 1\,d\theta=\pi.
$$
If you can show the two integrals are equal, then they each have to be $\pi/2$.
But they have to be equal since the graph of $\sin^2$ has the same size and shape as that of $\cos^2$ and the interval from $-\pi/2$ to $\pi/2$ is a full period.
A: If you look at the plot of $y=\cos^2x$, you will notice that the curve is perfectly symmetric around $y=\frac12$ and has period $\pi$, as confirmed by $\cos^2x=\frac12\cos2x+\frac12$.
For this reason, the average value over a period is $\frac12$, and the area under the curve is $\frac\pi2$.
A: Here's yet another way:
\begin{align}
& \int \cos^2\theta\,d\theta = \int(\cos\theta) \Big(\cos\theta\,d\theta\Big) = \int u\,dv = uv-\int v\,du \\[10pt]
= {} & -\cos\theta\sin\theta -\int(\sin\theta)\, \Big( -\sin\theta\, d\theta\Big) \\[10pt]
= {} & -\cos\theta\sin\theta + \int \sin^2\theta\,d\theta = -\cos^2\theta + \int(1-\cos^2\theta)\,d\theta \\[10pt]
= {} & -\cos\theta\sin\theta + \theta -\int\cos^2\theta\,d\theta.
\end{align}
So we have shown that
$$
\int \cos^2\theta\,d\theta = -\cos\theta\sin\theta + \theta -\int\cos^2\theta\,d\theta.
$$
Now add the same thing to both sides:
\begin{align}
& \phantom{{}+{}}\int \cos^2\theta\,d\theta = -\cos\theta\sin\theta  + \theta -\int\cos^2\theta\,d\theta. \\[12pt]
& {} + \int \cos^2\theta\,d\theta \phantom{{}= -\cos\theta\sin\theta + \theta }{} + \int\cos^2\theta\,d\theta \\[15pt]
& \phantom{+{}}2\int\cos^2\theta\,d\theta = -\cos\theta\sin\theta  + \theta + \text{constant}
\end{align}
and then divide both sides by $2$.
