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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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    You can use symmetry on the interval $[0, \pi/2]$: $\cos^2(\pi/2 - x) = \sin^2(x) = 1 - \cos^2(x)$.2012-06-01
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    That doesn't look very symmetric to me.2012-06-01
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    $x \mapsto \pi/2 - x$ reverses the interval. A picture should make clear what's happening.2012-06-01
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    I see someone did mention symmetry. I posted that as an answer, with details below.2012-06-01
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    @Jordan : Look at the graph of $y=\cos^2 x$ on the interval $0\le x\le \pi/2$. The graph of $y=\sin^2 x$ looks exactly the same except with left and right switched around. As $x$ goes from $0$ to $\pi/2$, the function $\cos^2 x$ behaves exactly the way the function $\sin^2 x$ behaves as $x$ goes in the opposite direction---from $\pi/2$ to $0$.2012-06-01
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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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    This formula comes from the fact that $\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$, like this: $\cos(2x) = \cos(x + x) = \cos(x)\cos(x) - \sin(x)\sin(x) = \cos^2(x) - \sin^2(x)$. So, from the fact that $\sin^2(x) = 1 - \cos^2(x)$, we have that: $\cos(2x) = \cos^2(x) - (1 - \cos^2(x)) = 2\cos^2(x) - 1$. Then: $\cos(2x) = 2\cos^2(x) - 1$.2012-06-01
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    @JohnEngbers I still can not get the correct answer even from that point. I am getting $1/2(-cos2x/4 + x)$2012-06-02
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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 $$ \int_0^{\pi/2} \cos^2 x\,dx + \int_0^{\pi/2} \sin^2 x\,dx = \int_0^{\pi/2} \left(\cos^2 x + \sin^2 x\right)\,dx = \int_0^{\pi/2} 1\,dx = \frac\pi 2. $$

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

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    I do no follow the last part, why are the two integrals equal to one? Is the only way I could do this problem is having memorized the graphs of all the trig squared fucntions? That seems incredibly absurd.2012-06-01
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    @Jordan $\cos^2 x + \sin^2 x = 1$ is a well known identity . . .and you should know *at least* the $\sin$ and $\cos$ curves since you passed trig.2012-06-01
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    @MichaelHardy I do not follow what is happening, did you make a mistake? Shouldn't the bounds be 0 to $\pi / 4$|2012-06-02
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    @Jordan : No. If it's $0$ to $\pi/4$, then you don't have that symmetry. The integrals of $\sin^2$ and $\cos^2 x$ over the integral from $0$ to $\pi/4$ are not equal to each other.2012-06-02
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    @MichaelHardy I do not understand why sin was introduced into this problem, well I get why but I do not understand why that is "legal" .2012-06-02
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    @Jordan : I confess I am a bit puzzled by your question. If one can demonstrate that for reasons of symmetry the two integrals are both the same, and one can find their sum, then it logically follows that each of them is half of that sum.2012-06-03
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    @MichaelHardy That makes even less sense to me. I guess we are thinking of this in terms of what the functions look like on a graph. They cross at some point and the area under that is the answer?2012-06-03
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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