$ \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx $

Question

$$ \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx $$

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is there more efficient and elegant way? Other than the solution below.

a solution:

Let $ I = \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx $,

$$ \int \frac{\sin^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx = \int \frac{\sin^{8}(x) + \cos^{8}(x) - \cos^{8}(x)}{\sin^{8}(x) + \cos^{8}(x)} dx = x - \int \frac{\cos^{8}(x)}{ \sin^{8}(x) + \cos^{8}(x) }dx $$

so we have $2I$ is:

$$ 2I = x + \int \frac{\sin^{8}(x)-\cos^{8}(x)}{ \sin^{8}(x) + \cos^{8}(x) }dx $$

Now,

$$ \sin^{8}(x) - \cos^{8}(x) = \left[ \sin^{4}(x) - \cos^{4}(x) \right] \left[ \sin^{4}(x) + \cos^{4}(x) \right] $$ $$ = \left[ \sin^{2}(x) - \cos^{2}(x) \right]\left[ \sin^{2}(x) + \cos^{2}(x) \right] \left[ (\sin^{2}(x) + \cos^{2}(x))^{2} - 2 \sin^{2}(x)\cos^{2}(x) \right] $$ $$ = \left[ \sin^{2}(x) - \cos^{2}(x) \right] \left[1 - 2 \sin^{2}(x)\cos^{2}(x) \right] $$

For the denominator:

$$ \sin^{8}(x) + \cos^{8}(x) = \left[ \sin^{4}(x) + \cos^{4}(x) \right]^{2} - 2 \sin^{4}(x)\cos^{4}(x) $$ $$ = \left[ (\sin^{2}(x) + \cos^{2}(x))^{2} - 2 (\sin(x)\cos(x))^{2} \right]^{2} - 2 \sin^{4}(x)\cos^{4}(x) $$ $$ = \left[1 - 2 (\sin(x)\cos(x))^{2} \right]^{2} - 2 (\sin(x)\cos(x))^{4} $$

Now

$$ 2I =x+ \int \frac{ \left[ \sin^{2}(x) - \cos^{2}(x) \right] \left[1 - 2 \sin^{2}(x)\cos^{2}(x) \right] }{\left[1 - 2 (\sin(x)\cos(x))^{2} \right]^{2} - 2 (\sin(x)\cos(x))^{4}} dx $$

Let $U = \sin(x) \cos(x)$, then $U'(x) = \cos^{2}(x) - \sin^{2}(x)$, so

$$ 2I = x- \int \frac{\left[1 - 2 U^{2} \right] }{\left[1 - 2 U^{2} \right]^{2} - 2 U^{4}} dU $$ $$ = x- \int \frac{\left[ 2 U^{2} - 1 \right] }{ (1 - (2 + \sqrt{2})U^{2})(1 + (\sqrt{2} - 2)U^{2})} dU $$

And from here I can continue using Partial Fraction: $$- \int \frac{\left[ 2 U^{2} - 1 \right] }{ (1 - (2 + \sqrt{2})U^{2})(1 + (\sqrt{2} - 2)U^{2})} dU = -\frac{1}{2}\int \frac{1}{ (1 - (2 + \sqrt{2})U^{2})} + \frac{1}{(1 - (2 - \sqrt{2})U^{2})} dU $$

and then for each of the 2 terms we use Partial Fraction again. First, let $\alpha = \sqrt{2+\sqrt{2}}U$, $\beta = \sqrt{2-\sqrt{2}}U$, so we have

$$ = -\frac{1}{2\sqrt{2+\sqrt{2}}}\int \frac{1}{ 1 - \alpha^{2}} d \alpha - \frac{1}{2\sqrt{2-\sqrt{2}}} \int \frac{1}{(1 - \beta^{2})} d \beta $$

then we have

$$ = -\frac{1}{2\sqrt{2+\sqrt{2}}} \left( \int \frac{1/2}{1-\alpha} d \alpha + \int \frac{1/2}{1+\alpha} d\alpha \right) - \frac{1}{2\sqrt{2-\sqrt{2}}} \left( \int \frac{1/2}{1-\beta } d \beta + \int \frac{1/2}{1 + \beta } d \beta \right) $$

$$ = -\frac{1}{4\sqrt{2+\sqrt{2}}} \left( -\ln(1-\alpha) + \ln(1+\alpha) \right) - \frac{1}{4\sqrt{2-\sqrt{2}}} \left( -\ln(1-\beta) + \ln(1+\beta) \right) $$

Answer 1

Another option would be to enforce the substitution $x=\frac {\pi}2-y$ and rewrite each sine and cosine term using their respective double angle identity.

The numerator is a direct application of

$$\cos 2y=2\cos^2y-1$$

Whereas the denominator can be expressed using a bit of algebraic manipulation

$$\begin{align*}\sin^8 y+\cos^8 y & =\left(\sin^4 y+\cos^4 y\right)^2-2\sin^4 y\cos^4 y\\ & =\left(1-\frac 12\sin^2 2y\right)^2-\frac 18\sin^4 2y\\ & =\color{red}{1-\sin^2 2y+\frac 18\sin^4 2y}\\ & =\color{blue}{\frac 18+\frac 34\cos^2 2y+\frac 18\cos^4 2y}\end{align*}$$

The colored expressions will be the relations used in the computation below $$\begin{align*}I & =-\int\frac {\cos^8 y}{\sin^8y+\cos^8y}\,\mathrm dy\\ & =-\frac 12\int\frac {(1+\cos 2y)^4}{1+6\cos^22y+\cos^42y}\,\mathrm dy\\ & =-\frac y2-2\int\frac {\cos 2y(2-\sin^22y)}{8-8\sin^22y+\sin^42y}\,\mathrm dy\\ & =-\frac y2-\int\frac {2-v^2}{8-8v^2+v^4}\,\mathrm dv\end{align*}$$

Where the substitution $v=\sin2y$ was made between the penultimate and the ultimate line. The remaining integral can be tackled in whatever way the reader sees fit.

Note that for the general case of the integral

$$J=\int\frac {\mathrm dv}{v^4+av^2+b^2}$$

The result can be obtained as such

$$\begin{align*}J & =\frac 12\int\frac {2+\frac b{v^2}-\frac b{v^2}}{v^2+\frac {b^2}{v^2}+a}\,\mathrm dv\\ & =\frac 12\int\frac {1+\frac b{v^2}}{\left(v-\frac bv\right)^2+a+2b}\,\mathrm dv+\frac 12\int\frac {1-\frac b{v^2}}{\left(v+\frac bv\right)^2+a-2b}\,\mathrm dv\\ & =\frac 12\int\frac {\mathrm dw_-}{w_-^2+a+2b}+\frac 12\int\frac {\mathrm dw_+}{w_+^2+a-2b}\end{align*}$$

Where the sign of the $a\pm 2b$ term determines whether the two resultant integrals can be expressed in terms of the arctangent function or natural logarithms.

Answer 2

Continue with

$$ 2I = x + \int \frac{\sin^{8}x-\cos^{8}x}{ \sin^{8}x + \cos^{8}x }dx =x+\int \frac{\tan^{8}x-1}{ \tan^{8}x + 1}dx \tag1$$ where, with $t = \tan x$ \begin{align} &\int \frac{\tan^{8}x-1}{ \tan^{8}x + 1}dx \\ =& \int \frac{(\tan^{4}x+1)(\tan^2x-1)\ \sec^2x}{ (\tan^{4}x -\sqrt2\tan^2x+1) (\tan^{4}x -\sqrt2\tan^2x+1)}dx \\ =&\ \frac12\int \frac{1-t^2}{1-\sqrt2t^2+t^4} + \frac{1-t^2}{1+\sqrt2t^2+t^4}\ dt\\ =& -\frac1{2\sqrt{2-\sqrt2}}\coth^{-1} \frac{t+\frac1t}{\sqrt{2-\sqrt2}} — \frac1{2\sqrt{2+\sqrt2}}\coth^{-1} \frac{t+\frac1t}{\sqrt{2+\sqrt2}} \end{align} Substitute into (1) to obtain $$ I = \frac x2 -\frac1{4\sqrt{2-\sqrt2}}\coth^{-1} \frac{\cot x+{\tan x}}{\sqrt{2-\sqrt2}}-\frac1{4\sqrt{2+\sqrt2}}\coth^{-1} \frac{\cot x+\tan x}{\sqrt{2+\sqrt2}}\\ $$

Answer 3

Instead, we are going to evaluate $$ \begin{aligned} A=\int \frac{\cos ^{8} x}{\sin ^{8} x+\cos ^{8} x} d x=\int \frac{1}{\tan ^{8} x+1} d x\stackrel{t=\tan x}{=} \int \frac{d t}{\left(1+t^{2}\right)\left(1+t^{8}\right)} \end{aligned} $$ Noting that $$ \frac{1}{(1+z )(1+z^{4})} = \frac{1}{2} \cdot \frac{1}{z+1}+\frac{1}{4} \cdot \frac{1-z}{z^{2}+\sqrt{2} z+1}+\frac{1}{4} \cdot\frac{1-z}{z^{2}-\sqrt{2} z+1} $$

Replacing $z$ by $t^2$ gives $$ \int\frac{d t}{\left(1+t^{2}\right)\left(1+t^{8}\right)}=\frac{1}{2} \underbrace{\int \frac{d t}{t^{2}+1}}_{x+C_1}+\frac{1}{4} \underbrace{\int \frac{1-t^{2}}{t^{4}+\sqrt{2} t^{2}+1}}_{J} d t+\frac{1}{4} \underbrace{\int \frac{1-t^{2}}{t^{4}-\sqrt{2} t^{2}+1}}_{k} d t $$ $$ \begin{aligned} J&=-\int \frac{1-\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}+\sqrt{2}} d t\\&=-\int \frac{d\left(t+\frac{1}{t}\right)}{\left(t+\frac{1}{t}\right)^{2}-(2-\sqrt{2})} \\&=-\frac{1}{2 \sqrt{2-\sqrt{2}}} \ln \left|\frac{t+\frac{1}{t}-\sqrt{2-\sqrt{2}}}{t+\frac{1}{t}+\sqrt{2-\sqrt{2}}}\right|+C_{2} \end{aligned} $$ Similarly, $$ K=-\frac{1}{2 \sqrt{2+\sqrt{2}}} \ln \left|\frac{t+\frac{1}{t}-\sqrt{2+\sqrt{2}}}{t+\frac{1}{t}+\sqrt{2+\sqrt{2}}}\right|+C_{3} $$ $$ \begin{aligned} A=& \frac{1}{2} x-\frac{1}{8 \sqrt{2-\sqrt{2}}} \ln \left|\frac{\tan ^{2} x-\sqrt{2-\sqrt{2}} \tan x+1}{\tan ^{2} x+\sqrt{2-\sqrt{2}} \tan x+1}\right| \\&-\frac{1}{8 \sqrt{2+\sqrt{2}}} \ln \left|\frac{\tan ^{2} x-\sqrt{2+\sqrt{2}} \tan x+1}{\tan ^{2} x+\sqrt{2+\sqrt{2}} \tan x+1}\right|+C_4 \end{aligned} $$

$$\begin{aligned}\int \frac{\sin ^{8} x}{\sin ^{8} x+\cos ^{8} x} dx=&x-A\\=& \frac{1}{2} x+\frac{1}{8 \sqrt{2-\sqrt{2}}} \ln \left|\frac{\tan ^{2} x-\sqrt{2-\sqrt{2}} \tan x+1}{\tan ^{2} x+\sqrt{2-\sqrt{2}} \tan x+1}\right|\\& +\frac{1}{8 \sqrt{2+\sqrt{2}}} \ln \left|\frac{\tan ^{2} x-\sqrt{2+\sqrt{2}} \tan x+1}{\tan ^{2} x+\sqrt{2+\sqrt{2}} \tan x+1}\right|+C \end{aligned}$$