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$$\int_{0}^{\pi/ n}{\sin^3(x)\over [1+\cos^2(x)]^2}\mathrm dx=F(n)\tag1$$

For $n\ge 1$

I have calculated out for:

$F(1)=1$

$F(2)={1\over 2}$

$F(3)={1\over 10}$

The general for $F(n)$ I wasn't able to find...

What is the closed form for $(1)?$

An attempt

Using $\cos^2(x)={1+\cos(2x)\over 2}$ then $1+\cos^2(x)={3+\cos(2x)\over 2}$

Calling upon a subsitition $u=3+\cos(2x)$ then $du=-2\sin(2x)dx$

Then $(1)$ becomes:

$$-\int_{1}^{\cos(2\pi/n)}{\sin^2(x)\over u^2\cos(x)}\mathrm du\tag2$$

$2+2\cos^2(x)=u$ then $\sin^2(x)={4-u\over 2}$

$\cos(x)=\sqrt{u-2\over 2}$

$${\sqrt{2}\over 2}\int_{1}^{\cos(\pi/n)}{u-4\over u^2\sqrt{u-2}}\mathrm du\tag3$$

$${\sqrt{2}\over 2}\int_{1}^{\cos{(\pi/n)}}{1\over u\sqrt{u-2}}\mathrm du+{2\sqrt{2}}\int_{1}^{\cos(\pi/n)}{1\over u^2\sqrt{u-2}}\mathrm du\tag4$$

From Standard integral table

$$\int{dx\over x\sqrt{ax+b}}={2\over \sqrt{-b}}\arctan\sqrt{ax+b\over -b}$$

$$\int{dx\over x^2\sqrt{ax+b}}=-{\sqrt{ax+b}\over bx}-{a\over 2b}\int{dx\over x\sqrt{ax+b}}$$

Substitute it in,

$${\sqrt{2}\over 2}\cdot{2\over\sqrt{2}}\arctan\sqrt{u-2\over 2}-2\sqrt{2}\left[-{\sqrt{u-2\over -2u}}+{1\over 4}\left({2\over\sqrt{2}}\cdot\arctan{\sqrt{u-2\over2}}\right)\right]\tag5$$

Break down to

$$\arctan\sqrt{u-2\over 2}-\sqrt{2}\cdot{\sqrt{u-2}\over u}-\arctan\sqrt{u-2\over 2}\tag6$$

$$=-\sqrt{2}\cdot{\sqrt{u-2}\over u}\tag7$$

I don't thing I am on the right track, any help?

2 Answers 2

5

Hint: \begin{align*} F(n)&=-\int_{0}^{\pi/ n}{\sin^2(x)\over [1+\cos^2(x)]^2}\mathrm d\cos x\\ &=-\int_{0}^{\pi/ n}{1-\cos^2(x)\over [1+\cos^2(x)]^2}\mathrm d\cos x\\\tag1 &=\int_{\cos\left ( \frac{\pi }{n} \right )}^{1}\frac{1-x^{2}}{\left ( 1+x^{2} \right )^{2}}\, \mathrm{d}x\\ &=\frac{1}{2}-\frac{\cos\left ( \dfrac{\pi }{n}\right )}{1+\cos^2\left ( \dfrac{\pi }{n}\right )} \end{align*} $(1):$ $\displaystyle \int \frac{1-x^{2}}{\left ( 1+x^{2} \right )^{2}}\, \mathrm{d}x=\frac{x}{1+x^{2}}+C$

0

$$ \int_{0}^{\pi/ n}{\sin^3(x)\over [1+\cos^2(x)]^2}\mathrm dx $$

$$ = \int_{0}^{\pi/n}\frac {(1-\cos^2 x)(\sin x) }{(1+\cos^2 x)^2} dx $$

Put $ \cos(x) = t$ , then, $\sin(x)dx = dt$

Rewriting the integral,

$$-\int_{1}^{\cos\frac{\pi}{n}} \frac{1-t^2}{(1+t^2)^2} $$

$$ = F(n) = \frac {1}{2}-\frac{\cos\frac {\pi}{n}}{\cos^2(\frac {\pi}{n})+1} $$

  • 1
    Hi and welcome to the site. As you see, there is another answer that is essentially the same as yours and that was accepted in January. I don't know about your motivation to add this answer. Also, a comment on the formulas: $cos(x)$ (`cos(x)`) looks better if you add a backslash in front of it, $\cos(x)$ (`\cos(x)`). The size of your final formula makes it look very bad. I hope I do not discourage you. Again, welcome to the site. – 2017-05-26
  • 0
    I understand, I'll make sure it doesn't happen again. Thanks :) – 2017-05-26