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This seems really simple but I can't get it $\int_0^{ \pi/2} \cos^2 x \,dx$

$u = \cos^ 2 x$, $du = -2 \cos x \sin x$

$dv = dx$, $v = x$

$x \cos x + 2 \int x \cos x \sin x$

$t = \sin x$, $dt = \cos x dx$

$2\int x \cos x t \, dt/ \cos x$

$2\int xt \, dt$

$2\int xt \, dt$

This is where I am stuck and I do not know what to do. I guess I can do the integration by parts again but it doesnt seem to help. I do not know if it is legal to work with two variables like that.

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    None of your integrals should have a mixture of $x$s and $t$s. That's just asking for trouble.2012-06-01

4 Answers 4

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This is one of those tricks to file away in your head (and no, you don't want 2 variables floating around in an integral like that). Utilize $\cos^2 x = \frac{1}{2} + \frac{\cos (2x)}{2},$ which is the standard half (or double?) angle formula from trig. After this initial substitution, you should be able to integrate.

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    @Jordan you should be integrating $1/2(\cos (2x) + 1)$ which will give you $1/2 (\sin (2x)/2 + x)$. Evaluating at $\pi/2$ and subtracting the value at $0$ gives $1/2 (0 + \pi/2) - 1/2 (0 + 0) = \pi/4$.2012-06-03
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You really don't need an antiderivative for this one if you use a simpler way to do it. Notice that $ \int_0^{\pi/2} \cos^2 x\,dx $ must be the same as $ \int_0^{\pi/2} \sin^2 x\,dx $ because both graphs have the same size and shape; one of them is a mirror-image of the other, with the "mirror" at $x=\pi/4$.

Then notice that

\begin{align} & \int_0^{\pi/2} \cos^2 x\,dx + \int_0^{\pi/2} \sin^2 x\,dx \\[8pt] = {} & \int_0^{\pi/2} \left(\cos^2 x + \sin^2 x\right)\,dx \\[8pt] = {} & \int_0^{\pi/2} 1\,dx = \frac\pi 2. \end{align}

Therefore either integral separately is $\pi/4$.

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    Go to Wolfram alpha and enter this command: plot y = cos^2(x), x=0 to \pi/2. Then try plot y = sin^2(x), x=0 to \pi/2. The two graphs have the same size and shape; hence the areas under them are the same. And the area under the _sum_ of the two functions is just the area of a rectangle, so it's easy to find. You're certainly _not_ looking for the area under the point where they cross; I don't know why you would say that.2012-06-03
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Try the reduction formula I showed in the answer to your question.

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An overkill. Since the Beta function can be written as $\int_{0}^{\pi/2}\sin^{m}\left(x\right)\cos^{n}\left(x\right)dx=\frac{B\left(\frac{n+1}{2},\frac{m+1}{2}\right)}{2}$ we have $\int_{0}^{\pi/2}\cos^{2}\left(x\right)dx=\frac{B\left(\frac{1}{2},\frac{3}{2}\right)}{2}=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{3}{2}\right)}{2\Gamma\left(2\right)}=\color{red}{\frac{\pi}{4}}.$

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    I'm not particularly familiar with the Beta function but nice answer. Welcome change!2017-01-09