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I want to evaluate the following definite interal \begin{gather*} \int_{0}^{2m\pi} \frac{1}{\sin^4(x)+\cos^4(x)}d x, \end{gather*} where $m$ is a given positive integer. I have calculated this integral. But my method is fairly complex.

My method is below. First, observe that \begin{align*} &\quad \sin^4(x)+\cos^4(x)=\left(\sin^2(x)+\cos^2(x)\right)^2-2\sin^2(x)\cos^2(x)\\ &=1-\frac{1}{2}\sin^2(2x)=1-\frac{1}{2}\left(\frac{1-\cos(4x)}{2}\right)=\frac{\cos(4x)}{4}+\frac{3}{4}, \end{align*} and the function $x\mapsto \cos(4x)$has period $\frac{2\pi}{4}=\frac{\pi}{2},$ thus the fuction $x\mapsto \sin^4(x)+\cos^4(x)$ has also period $\frac{\pi}{2}.$ As a result, the period of the integrand $x\mapsto \frac{1}{\sin^4(x)+\cos^4(x)}$ is $\frac{\pi}{2}.$ Consequently, \begin{align*} &\quad \int_{0}^{2m\pi}\frac{1}{\sin^4(x)+\cos^4(x)}d x=\sum_{j=0}^{4m-1}\int_{j\frac{\pi}{2}}^{(j+1)\frac{\pi}{2}}\frac{1}{\sin^4(x)+\cos^4(x)}d x\\ &=\sum_{j=0}^{4m-1}\int_{j\frac{\pi}{2}}^{j\frac{\pi}{2}+\frac{\pi}{2}}\frac{1}{\sin^4(x)+\cos^4(x)}d x=\sum_{j=0}^{4m-1}\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin^4(x)+\cos^4(x)}d x\\ &=4m\int_{0}^{\frac{\pi}{2}}\frac{1}{\sin^4(x)+\cos^4(x)}d x=4m\int_{-\frac{\pi}{4}}^{-\frac{\pi}{4}+\frac{\pi}{2}}\frac{1}{\sin^4(x)+\cos^4(x)}d x\\ &=4m\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\frac{1}{\sin^4(x)+\cos^4(x)}d x=8m\int_{0}^{\frac{\pi}{4}}\frac{1}{\sin^4(x)+\cos^4(x)}d x.\tag{26.1} \end{align*} By using trigonometric substitution, \begin{align*} & \quad \int_{0}^{\frac{\pi}{4}}\frac{1}{\sin^4(x)+\cos^4(x)}d x=\int_{0}^{\frac{\pi}{4}}\frac{\sec^2(x)}{1+\tan^4(x)}\cdot\sec^2(x)d x\\ &=\int_{0}^{\frac{\pi}{4}}\frac{1+\tan^2(x)}{1+\tan^4(x)}d \tan(x)=\int_{0}^{1}\frac{1+y^2}{1+y^4}d y\qquad ({y=\tan(x)})\\ &=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{y^2+\sqrt{2}y+1}+\frac{1}{y^2-\sqrt{2}y+1}\right)d y\\ &=\frac{1}{2}\int_{0}^{1}\left(\frac{1}{\left(y+\frac{\sqrt{2}}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2}+\frac{1}{\left(y-\frac{\sqrt{2}}{2}\right)^2+\left(\frac{\sqrt{2}}{2}\right)^2}\right)d y\\ &=\frac{1}{\sqrt{2}}\big(\arctan(\sqrt{2}y+1)+\arctan(\sqrt{2}y-1)\big)\bigg|_{0}^1\\ &=\frac{1}{\sqrt{2}}\left(\arctan(\sqrt{2}+1)+\arctan(\sqrt{2}-1)\right)\\ &=\frac{1}{\sqrt{2}}\left(\arctan(\sqrt{2}+1)+\arctan\left(\frac{1}{\sqrt{2}+1}\right)\right)\\&=\frac{1}{\sqrt{2}}\cdot\frac{\pi}{2}=\frac{\pi}{2\sqrt{2}}.\tag{26.2} \end{align*} Finally, inserting (26.2) into (26.1), we arrive at \begin{align*} \int_{0}^{2m\pi}\frac{1}{\sin^4(x)+\cos^4(x)}d x=8m\cdot \frac{\pi}{2\sqrt{2}}=2\sqrt{2}\,m\pi. \end{align*}

My question is: Is there any simple method to evaluate this definite integral?

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  • $\begingroup$ Have you tried $u = \tan \frac x2$? $\endgroup$ – MathematicsStudent1122 Dec 6 '16 at 1:59
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    $\begingroup$ @MathematicsStudent1122: Since the half-angle substitution is more complex than the whole-angle substitution, I used $y=\tan(x).$ $\endgroup$ – azc Dec 6 '16 at 2:11
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HINT.-$$(\cos^2(x)+\sin^2(x))^2=1\Rightarrow\cos^4(x)+\sin^4(x)=1-\dfrac{\sin^2(2x)}{2}$$ and $$\int\dfrac{2dx}{2-\sin^2(2x)}=\frac{\arctan(\frac{\tan(2x)}{\sqrt2}}{\sqrt2}+C$$

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  • $\begingroup$ I've got it. Your method is easy! $\endgroup$ – azc Dec 7 '16 at 4:50
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Hint:

$$\frac1{\cos^4x+\sin^4x}=\frac{1+\tan^2x}{1+\tan^4x}\sec^2x$$

Set $\tan x=y$

$$\frac{1+y^2}{1+y^4}=\frac{1/y^2+1}{(y-1/y)^2+2}$$

Set $y-1/y=z$

So directly $2z=-\cot2x$

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  • $\begingroup$ Since $0\leq x\leq \pi/4,$ there is a singularity at $x=0,$ if set $y-1/y=z,$ where $\tan(x)=y.$ Thus I think the substitution $y-1/y=z$ is not so good. $\endgroup$ – azc Dec 7 '16 at 4:24
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First, by periodicity $$\int_{0}^{2m\pi} \frac{1}{\sin^4(x)+\cos^4(x)}d x,=m\int_{0}^{2\pi} \frac{1}{\sin^4(x)+\cos^4(x)}d x=m\int_{0}^{2\pi} \frac{4}{3+\cos(4x)}d x\\ =m\int_{0}^{8\pi} \frac{1}{3+\cos(u)}d u=4m\int_{0}^{2\pi} \frac{1}{3+\cos(u)}d u$$

This integral is easy to evaluate by complex analysis: $$\int_{0}^{2\pi} \frac{1}{3+\cos(u)}d u= \int_{|z|=1} \frac{1}{3+\frac{z+\frac{1}{z}}{2}}\frac{dz}{iz}\\ =\frac{1}{i}\int_{|z|=1} \frac{1}{z^2+6z+1} dz =2 \pi Res(f,2\sqrt{2}-3) \\ =2 \pi \frac{1}{4\sqrt{2}-6+6}$$

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  • $\begingroup$ But the problem is from a textbook of elementary calculus course, where the classmates are not knowing anything about Complex Analysis. $\endgroup$ – azc Dec 7 '16 at 4:25

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