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I have tried Trigonometric Substitution, but I can´t get an already known function to be easy for integrate:

$$\int\frac{dx}{1+x^2+\sin^2x}$$

I entered this on Wolfram and it gave me the same function. I'm not asking for the exact solution, just a good way to solve it.

Regards.

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    $\begingroup$ I'll bet you a nickel that there's a $dx$ in their somewhere. $\endgroup$
    – John
    Commented Feb 11, 2014 at 20:49
  • $\begingroup$ I'm sorry, yes it is. $\endgroup$
    – ismatim
    Commented Feb 11, 2014 at 20:50
  • $\begingroup$ @John. In "their" or "there" :) $\endgroup$
    – imranfat
    Commented Feb 11, 2014 at 20:54
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    $\begingroup$ Sorry about the question, but we use different notation. Is it $\sin x^2$ or $\sin^2 x$? $\endgroup$
    – Vadim
    Commented Feb 11, 2014 at 20:54
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    $\begingroup$ It is highly possible that an exact antiderivative might not exist without using special functions. $\endgroup$
    – robjohn
    Commented Feb 11, 2014 at 21:13

1 Answer 1

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Hint:

$\because0\leq\dfrac{\sin^2x}{x^2+1}<1$ $\forall x\in\mathbb{R}$

$\therefore\int\dfrac{dx}{1+x^2+\sin^2x}$

$=\int\dfrac{dx}{(x^2+1)\left(1+\dfrac{\sin^2x}{x^2+1}\right)}$

$=\int\dfrac{1}{x^2+1}\sum\limits_{m=0}^\infty\dfrac{(-1)^m\sin^{2m}x}{(x^2+1)^m}dx$

$=\int\dfrac{1}{x^2+1}dx+\int\sum\limits_{m=1}^\infty\dfrac{(-1)^m\sin^{2m}x}{(x^2+1)^{m+1}}dx$

$=\int\dfrac{1}{x^2+1}dx+\int\sum\limits_{m=1}^\infty\dfrac{(-1)^m(2m)!}{4^m(m!)^2(x^2+1)^{m+1}}dx+\int\sum\limits_{m=1}^\infty\sum\limits_{n=1}^m\dfrac{(-1)^{m+n}(2m)!\cos2nx}{2^{2m-1}(m-n)!(m+n)!(x^2+1)^{m+1}}dx$

$=\int\dfrac{1}{x^2+1}dx+\int\sum\limits_{m=1}^\infty\dfrac{(-1)^m(2m)!}{4^m(m!)^2(x^2+1)^{m+1}}dx+\int\sum\limits_{m=1}^\infty\sum\limits_{n=1}^m\sum\limits_{p=0}^\infty\dfrac{(-1)^{m+n+p}(2m)!4^pn^{2p}x^{2p}}{2^{2m-1}(m-n)!(m+n)!(2p)!(x^2+1)^{m+1}}dx$

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