# int_0^pi(xsin^4x)/(sin^4x+cos^4x)dx = ?

May 3, 2018

${\pi}^{2} / 4$.

#### Explanation:

We will use these Results :

$\left({R}_{1}\right) : {\int}_{0}^{a} f \left(x\right) \mathrm{dx} = {\int}_{0}^{a} f \left(a - x\right) \mathrm{dx}$.

$\left({R}_{2}\right) \left(i\right) : {\int}_{0}^{2 a} f \left(x\right) \mathrm{dx} = 0 , \mathmr{if} f \left(2 a - x\right) = - f \left(x\right) , \mathmr{and} ,$

$\left({R}_{2}\right) \left(i i\right) : {\int}_{0}^{2 a} f \left(x\right) \mathrm{dx} = 2 {\int}_{0}^{a} f \left(x\right) \mathrm{dx} , \mathmr{if} f \left(2 a - x\right) = f \left(x\right)$.

Let, $I = {\int}_{0}^{\pi} \frac{x {\sin}^{4} x}{{\sin}^{4} x + {\cos}^{4} x} \mathrm{dx}$.

$\therefore I = {\int}_{0}^{\pi} \frac{\left(\pi - x\right) {\sin}^{4} \left(\pi - x\right)}{{\sin}^{4} \left(\pi - x\right) + {\cos}^{4} \left(\pi - x\right)} \mathrm{dx} , \ldots \left[\left({R}_{1}\right)\right]$

$= {\int}_{0}^{\pi} \frac{\left(\pi - x\right) {\sin}^{4} x}{{\sin}^{4} x + {\cos}^{4} x} \mathrm{dx}$,

$= \pi {\int}_{0}^{\pi} {\sin}^{4} \frac{x}{{\sin}^{4} x + {\cos}^{4} x} \mathrm{dx} - {\int}_{0}^{\pi} \frac{x {\sin}^{4} x}{{\sin}^{4} x + {\cos}^{4} x} \mathrm{dx}$,

$i . e . , I = \pi {\int}_{0}^{\pi} {\sin}^{4} \frac{x}{{\sin}^{4} x + {\cos}^{4} x} \mathrm{dx} - I$,

$\mathmr{and} , 2 I = \pi {\int}_{0}^{\pi} {\sin}^{4} \frac{x}{{\sin}^{4} x + {\cos}^{4} x} \mathrm{dx}$.

Now, before applying $\left({R}_{2}\right) ,$ let us work out

$f \left(2 a - x\right) = f \left(\pi - x\right) \text{ for } f \left(x\right) = {\sin}^{4} \frac{x}{{\sin}^{4} x + {\cos}^{4} x}$.

From above, we see that, $f \left(\pi - x\right) = f \left(x\right)$.

Hence, by $\left({R}_{2}\right) \left(i i\right)$,

$\frac{2 I}{\pi} = 2 {\int}_{0}^{\frac{\pi}{2}} {\sin}^{4} \frac{x}{{\sin}^{4} x + {\cos}^{4} x} \mathrm{dx}$,

$\mathmr{and} , \frac{I}{\pi} = {\int}_{0}^{\frac{\pi}{2}} {\sin}^{4} \frac{x}{{\sin}^{4} x + {\cos}^{4} x} \mathrm{dx} \ldots \left(1\right)$.

But for $f \left(x\right) = {\sin}^{4} \frac{x}{{\sin}^{4} x + {\cos}^{4} x}$,

$f \left(\frac{\pi}{2} - x\right) = {\sin}^{4} \frac{\frac{\pi}{2} - x}{{\sin}^{4} \left(\frac{\pi}{2} - x\right) + {\cos}^{4} \left(\frac{\pi}{2} - z\right)} \mathrm{dx}$,

$= {\cos}^{4} \frac{x}{{\cos}^{4} x + {\sin}^{4} x}$.

Therefore, by applying $\left({R}_{1}\right)$, on $\frac{I}{\pi}$,

$\frac{I}{\pi} = {\int}_{0}^{\frac{\pi}{2}} {\cos}^{4} \frac{x}{{\cos}^{4} x + {\sin}^{4} x} \mathrm{dx} \ldots \ldots \left(2\right)$.

Finally, we add $\left(1\right) \mathmr{and} \left(2\right)$ to get,

$\frac{2 I}{\pi} = {\int}_{0}^{\frac{\pi}{2}} \frac{\left({\sin}^{4} x + {\cos}^{4} x\right)}{\left({\cos}^{4} x + {\sin}^{4} x\right)} \mathrm{dx}$,

$= {\int}_{0}^{\frac{\pi}{2}} 1 \mathrm{dx}$,

$= {\left[x\right]}_{0}^{\frac{\pi}{2}}$,

$\therefore \frac{2 I}{\pi} = \frac{\pi}{2} , \text{ giving, }$

$I = {\pi}^{2} / 4$.

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