How to evaluate $\int_0^{\pi/2}\;3\;\cos^2\theta\;d\theta$ $$\int_0^{\pi/2}\color{red}{3}\;\cos^2\theta\;d\theta$$
I tried doing it using $u$-substitution and then getting the anti-derivative. All in all, I keep getting $-1$, which is apparently wrong. I'd appreciate any help.
 A: There is a cute trick for this:
$$I = 3\int_{0}^{\pi/2} \cos^2 x \text{ dx} = 3\int_{0}^{\pi/2} \sin^2 \text{ dx}$$
by using $\int_{a}^{b} f(x) \text{ dx} = \int_{a}^{b} f(a+b-x) \text{ dx}$
Adding gives us
$$2I = 3 \int_{0}^{\pi/2} (\cos^2 x + \sin^2 x) \text{ dx} = 3 \int_{0}^{\pi/2} 1 \text{ dx} = \frac{3\pi}{2}$$
Hence your integral is
$$ \frac{3\pi}{4}$$
But, there are more general methods which can be used. See the answers here: Evaluating $\int P(\sin x, \cos x) \text{d}x$
A: Hint: Use the double-angle formula $\cos 2\theta=2\cos^2\theta-1$.  So $\cos^2\theta
=\frac{\cos 2\theta +1}{2}$.
I believe you mean to ask how does one evaluate, or compute the definite integral.
A: $$\int_0^\frac{\pi}{2} 3 \cos^2\theta d\theta= \int_0^\frac{\pi}{2} \frac{3}{2}(1+\cos2\theta)d\theta$$
$$=\int_0^\frac{\pi}{2} \frac{3}{2}+\frac{3}{2}\cos 2\theta d\theta$$
$$=\frac{3}{2}(\frac{\pi}{2}-0)+\frac{3}{4}(\sin2\frac{\pi}{2}-\sin 0)$$
$$=\frac{3}{4}$$
A: Just for fun, you can use integration by parts to find the antiderivative of $\cos^2 x$:
$$\eqalign{
\color{maroon}{\int\cos^2 x\,dx} =\int\underbrace{\cos x}_u\,\underbrace{\cos x\,dx}_{dv}&= 
 \underbrace{\cos x\vphantom{)}}_u\,\underbrace{( \sin x )}_{ v}-
\int\underbrace{\sin x\vphantom{)}}_v\,\underbrace{(-\sin x\,dx)}_{du}\cr
&=\sin x\cos x+\int\sin^2 x\,dx\cr
&=\sin x\cos x+\int(1-\cos^2 x)\,dx\cr
&=\sin x\cos x+ x-\color{maroon}{\int\cos^2 x \,dx};\cr
}
$$
Whence
$$
\int\cos^2 x\,dx={\sin x\cos x+x\over 2}+C.
$$
A: I would suggest Wallis' formula
$\int^{\frac{\pi}{2}}_0 \cos^n{x}$ $dx=\int^{\frac{\pi}{2}}_0 \cos^n{x}$ $dx$
For $n$ even:
$\int^{\frac{\pi}{2}}_0 \cos^n{x}$ $dx=\int^{\frac{\pi}{2}}_0 \cos^n{x}$ $dx=\frac{1}{2}$x$\frac{3}{4}$x$...\frac{(n-1)}{n}$x$\frac{\pi}{2}$
For $n$ odd:
$\int^{\frac{\pi}{2}}_0 \cos^n{x}$ $dx=\int^{\frac{\pi}{2}}_0 \cos^n{x}$ $dx=\frac{1}{2}$x$\frac{3}{4}$x$...\frac{(n-1)}{n}$
$\therefore$ from this formula,
$3\int^{\frac{\pi}{2}}_0 \cos^2{x}$ $dx=3(\frac{\pi}{4})$
