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Here it is:

$\int_o^\pi x\cos^4x\,dx$ I used integration by parts but I would be grateful if someone told me an alternate method to compute the integral faster.

4 Answers 4

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One has $\cos^4 x={1\over16}(e^{ix}+e^{-ix})={1\over8}\bigl(3+4\cos(2x)+\cos(4x)\bigr)$ and therefore $\eqalign{J&:=\int_0^\pi x\cos^4 x\ dx \cr &={1\over8}\Bigl(x(3x+2\sin(2x)+{1\over4}\sin(4x)\Bigr)\biggr|_0^\pi -{1\over8}\int_0^\pi \bigl(3x+2\sin(2x)+{1\over4}\sin(4x)\bigr)\ dx\ .\cr}$ Inspecting the right hand side we see that all that remains is $J={1\over8}\cdot 3\pi^2-{1\over8}\int_0^\pi 3x\ dx= {3\pi^2\over16}\ .$

6

Let $I=\int_0^\pi x\cos^4x\,dx$

Using $\int_a^b f(x)\,dx=\int_{a}^{b}f(a+b-x)\,dx$ (Proof),

$I=\int_0^\pi (\pi-x)\cos^4(\pi-x)\,dx$ as $\cos(\pi-x)=-\cos x$

$I=\pi\int_0^\pi \cos^4x\,dx -I $

$2I=\pi\int_0^\pi \cos^4x\,dx $

Now, if $f(x)=\cos^4x,f(-x)=\{\cos(-x)\}^4=\cos^4x=f(x)$ i.e., is an even function, so, $\int_0^\pi \cos^4x\,dx =2\int_0^{\frac\pi 2}\cos^4x\,dx$

Now, we can use reduction formula to find $\int_0^{\frac\pi 2}\cos^4x\,dx=\frac {4-1} 4 \frac {4-3} {4-2}\frac \pi 2$

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    This seems to be the fastest, of course, yet it relies on complex functions which the OP is, perhaps, not acquainted.2012-10-25
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The well known formula $\int_0^\pi xf(\sin x )dx=\frac \pi2\int_0^\pi f(\sin x )dx$ yields $I=\int_{o}^{\pi} x\cos^4x=\frac \pi2\int_0^\pi \cos^4xdx=\pi\int_0^{\pi/2} \cos^4xdx$ then let $x=\pi/2-y$ and have that $I=\pi\int_0^{\pi/2} \cos^4xdx=\pi\int_0^{\pi/2} \sin^4xdx$ $2I=\pi\int_0^{\pi/2} (\sin^4x+\cos^4x)dx=\frac \pi4\int_0^{\pi/2} (\cos(4 x)+3) dx= \frac{3 \pi^2}{8}$ $I= \frac{3 \pi^2}{16}.$

Chris

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    @Christian$B$latter: a nice interesting proof (+1)2012-10-25
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I'm not sure if this is "faster" than what you did but it's a different way to integrate $\,\cos^4x\,$. First,

$\cos^4x=\cos^2x(1-\sin^2x)=\cos^2x-\cos^2x\sin^2x$

$\int\cos^2x\,dx=\frac{x+\cos x\sin x}{2}\,.\,.\,.\,.\,(\text{Hint: use, for example, }\,\,\cos 2x=2\cos^2x-1):$

$\int\cos^2x\sin^2x\,dx=-\frac{\cos^3x\sin x}{3}+\frac{1}{3}\int\cos^4x\,dx\,.$

Thus, putting $\,K:=\int \cos^4x\,dx$ , we get:

$K=\frac{1}{2}(x+\cos x\sin x)+\frac{1}{3}(\cos^3x\sin x)-\frac{1}{3}K\Longrightarrow$

$K=\left.\frac{3}{8}\left(x+\cos x\sin x\right)+\frac{1}{4}\left(\cos^3x\sin x\right)\right|_0^\pi=\frac{3\pi}{8}$

and since $\,\displaystyle{I=\frac{\pi}{2}K}\,$ , by lab's answer, we're done.