# Let $f(x) = \int \frac{x}{1-x^{8}}dx\,$

1. Let $f(x) = \int \frac{x}{1-x^{8}}dx\,$.
• Represent $I(x)$ by a power series $\sum^{\infty}a_{n}x^{n}$.(Find $a_{n}$)
• What is the radius of convergence of $I(x)$ ?
2. Two curves are generated by polar equations $r=1+\sin\theta$ and $r=-\sin\theta$.
• Find the area of the region that lies inside both two curves.
• Find the length of the part of the curve $r=1+\sin\theta$ that lies inside the curve $r=-\sin\theta$.
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@user1439040 this must be two separate questions – Norbert Jul 20 '13 at 9:51

Hint: For $(1)$ you need the identity

$$\frac{1}{1-t}=\sum_{k=0}^{\infty} t^k.$$

$$f(x)= \int\frac{x}{1-x^8} dx = \int \sum_{k=0}^{\infty}x^{8k+1}dx = \sum_{k=0}^{\infty}\frac{x^{8k+2}}{8k+2} + c.$$

Now, try to find the radius of convergence using some standard techniques.

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HINT:

For the first problem, $$\frac x{1-x^8}=\frac x2\left(\frac1{1-x^4}+\frac1{1+x^4}\right)$$

$$=\frac x4\left(\frac1{1-x^2}+\frac1{1+x^2}\right)+\frac12\frac x{1+x^4}$$

$$=\frac x4\left(\frac1{1-x^2}\right)+\frac14\frac x{1+x^2}+\frac12\frac x{1+x^4}$$

$$=\frac 18\left(\frac{1+x-(1-x)}{1-x^2}\right)+\frac14\frac x{1+x^2}+\frac12\frac x{1+x^4}$$

$$=\frac 18\frac1{1-x}-\frac18\frac1{1+x}+\frac14\frac x{1+x^2}+\frac12\frac x{1+x^4}$$

For the last two terms put $x^2=u$

You will find all required the Series Formula here

In fact, we can make the substitution $x^2=u$ from the very start, but then we need substitute back $u$ with $x$ as we need the Power Series of $x$

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