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Integrate $\displaystyle \int \limits_0^\pi{{x\sin x}\over{1+\cos^2x}}dx$. I tried substituting $t=\cos x$, and then integrate with integration by parts. It got all messy... Thanks in advance for any help!

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Integration by parts is worth a try: $$ \begin{align} \int_0^\pi\frac{x\sin(x)}{1+\cos^2(x)}\,\mathrm{d}x &=-\int_0^\pi\frac{x}{1+\cos^2(x)}\,\mathrm{d}\cos(x)\\ &=-\int_0^\pi x\,\mathrm{d}\arctan(\cos(x))\\ &=-\pi\left(-\frac\pi4\right)+\int_0^\pi\arctan(\cos(x))\,\mathrm{d}x\\ &=\frac{\pi^2}{4}+\int_{-\pi/2}^{\pi/2}\arctan(\sin(x))\,\mathrm{d}x\\ &=\frac{\pi^2}{4} \end{align} $$ Since $\cos(x)=\sin(\pi/2-x)$ and then $\arctan(\sin(x))$ is an odd function.

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Let

$I =\displaystyle \int \limits_0^\pi{{x\sin x}\over{1+\cos^2x}}dx$.

Substitute $t=\pi - x$.

Note that sin($\pi - x$) = sin($x$), and cos($\pi - x$) = -cos($x$).

You will notice that;

$I = \text{"something to integrate using your original substitution"} - I$

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If you use this substitution $u=x-\frac{\pi}{2}$, it is much easier to get the answer. In fact \begin{eqnarray*} \int_0^\pi\frac{x\sin x}{1+\cos^2x}dx&=&\int_{-\pi/2}^{\pi/2}\frac{(u+\frac{\pi}{2})\cos u}{1+\sin^2u}dx\\ &=&\int_{-\pi/2}^{\pi/2}\frac{u\cos u}{1+\sin^2u}du+\frac{\pi}{2}\int_{-\pi/2}^{\pi/2}\frac{\cos u}{1+\sin^2u}du\\ &=&\frac{\pi}{2}\int_{-\pi/2}^{\pi/2}\frac{d\sin u}{1+\sin^2u}\\ &=&\frac{\pi}{2}\int_{-\pi/2}^{\pi/2}\frac{d\sin u}{1+\sin^2u}\\ &=&\frac{\pi}{2}\arctan(\sin u)\big|_{-\pi/2}^{\pi/2}\\ &=&\frac{\pi^2}{4}. \end{eqnarray*} Here we use the facts that if $f(u)$ is an odd function in $(-a,a)$, then $$ \int_{-a}^af(u)du=0. $$ Clearly $\frac{u\cos u}{1+\sin^2u}$ is an odd function.

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  • $\begingroup$ This is pretty close to my answer with the order of things switched around. (+1) $\endgroup$ – robjohn Apr 21 '13 at 15:04
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Hint: $\int_0^af(x)dx=\int_0^af(a-x)dx$ Use this to simplify it and then do what you were trying.

Edit: ohad asked why $\int_0^af(x)dx=\int_0^af(a-x)dx$ is true. This follows from making the substitution $t=a-x$, So, $\int_0^af(x)dx=-\int_a^0f(a-t)dt=\int_0^af(a-x)dx$. One can also look at it geometrically, by taking the same area under the curve and starting from a instead of 0.

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  • $\begingroup$ Why is that true? $\endgroup$ – ohad Apr 21 '13 at 10:30
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    $\begingroup$ @ohad Make a substitution t=a-x $\endgroup$ – Ishan Banerjee Apr 21 '13 at 10:33
  • $\begingroup$ No I meant, why is $\int_0^af(x)dx=\int_0^af(a-x)dx$ true? I just never seen it. Is it by some theorem? $\endgroup$ – ohad Apr 21 '13 at 10:37
  • $\begingroup$ @ohad: just as Ishan said: substitute $x=a-t$ to get $\int_0^af(x)\,\mathrm{d}x=\int_0^af(a-t)\,\mathrm{d}t$. Then, since $t$ is just a dummy variable, you get $\int_0^af(x)\,\mathrm{d}x=\int_0^af(a-x)\,\mathrm{d}x$. $\endgroup$ – robjohn Apr 21 '13 at 15:00
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Ohad, you can actually prove it. We want to prove $$\int_{0}^{a} f(x) \ dx = \int_{0}^{a} f(a-x) \ dx$$

To do this, do the following:

  • Put $t= a-x$. Then you actually have $dx= -dt$.

  • Your new integral then becomes $\displaystyle\int_{a}^{0} -f(a-t) \ dt$

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    $\begingroup$ This looks as if it was intended to be a comment to Ishan Banerjee's answer. It doesn't really answer the question. $\endgroup$ – robjohn Apr 21 '13 at 15:07

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