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Knowing that $f(x)f(-x)=1$ for all $x$, evaluate the following integral:

$$\int_{-\pi\over4}^{\pi\over4} {1\over{(1+2\sin^2x)(1+f(x))}} $$

Also, I found similar integrals of the form $\int_{-a}^{a} {1\over{(1+g(x))(1+f(x))}}$ where $g$ is even and $f$ satisfies the property above. I wonder if there's a nice trick for solving all integrals of that type, using general properties like the one saying that the integral of an odd function between $-a$ and $a$ is 0.

Thank you in advance for any solutions or hints.

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    $\begingroup$ Hint: $$\frac{1}{1+f(x)} + \frac{1}{1+f(-x)} = 1$$ $\endgroup$ – achille hui Jan 24 '14 at 13:44
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Hint: Start by noting that the substitution $w=-x$, $dw=-dx$ yields $$ \int_{-\pi/4}^{0}\frac{1}{(1+2\sin^2x)(1+f(x))}\,dx=\int_0^{\pi/4}\frac{1}{(1+2\sin^2w)(1+\frac{1}{f(w)})}\,dw. $$ Then try to combine this with the integral on $[0,\frac{\pi}{4}]$ as a single integral.

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As $$\int_a^bg(x)dx=\int_a^bg(a+b-x)dx,$$

if $a=-\dfrac\pi4,b=\dfrac\pi4, a+b-x=-x$

and $\displaystyle g(x)=\frac1{(1+2\sin^2x)[1+f(x)]}$

$\displaystyle g\left(-\dfrac\pi4+\dfrac\pi4-x\right)=\frac1{\left([1+2\sin^2\left(-x\right)\right]\left([1+f\left(-x\right)\right]}=\frac1{(1+2\sin^2x)[1+f(-x)]}$

$\displaystyle=\frac1{(1+2\sin^2x)\left[1+\dfrac1{f(x)}\right]}=\frac{f(x)}{(1+2\sin^2x)[1+f(x)]}$

Now,

$$I=\int_a^b g(x)dx=\int_a^b g(a+b-x)dx$$

$$\implies2I=\int_a^b g(x)dx+\int_a^b g(a+b-x)dx=\int_a^b\left[g(x)+g(a+b-x)\right]dx$$

Assumption: $f(x)+1\ne0$ as the integral will be undefined

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  • $\begingroup$ @Andreea, how about this? $\endgroup$ – lab bhattacharjee Jan 24 '14 at 15:37

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