How to calculate $$ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx? $$

I already know one possible way, that is by : $$ \int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx = \int 1 - \frac{\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx $$ $$= x- \int \frac{1}{1+\tan^{6}(x)} dx $$ Then letting $u=\tan(x)$, we must solve $$\int \frac{1}{(1+u^{6})(1+u^{2})} du $$ We can reduce the denominator and solve it using Partial Fraction technique. This is quite tedious, I wonder if there is a better approach.

Using same approach, for simpler problem, I get $$\int \frac{\sin^{3}(x)}{\sin^{3}(x)+\cos^{3}(x)} dx = \frac{x}{2} - \frac{\ln(1+\tan(x))}{6} + \frac{\ln(\tan^{2}(x)- \tan(x)+1)}{3} - \frac{\ln(\sec(x))}{2} + C$$

  • $\begingroup$ Actually partial fractions are not that difficult to figure with $1+x^6 = (1+x^2)(x^4-x^2+1)$ .. $\endgroup$ – King Tut Mar 1 '18 at 10:41
  • $\begingroup$ $$\int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx=\int\frac{\tan^6(x)}{1+\tan^6(x)} dx=\frac{1}{18}(9x-3\sin(x)\cos(x))+\sqrt3\log\left(\frac{\sqrt3-3\sin(x)\cos(x)}{\sqrt3+3\sin(x)\cos(x)}\right)+C$$ $\endgroup$ – Piquito Mar 1 '18 at 12:34

Let us take: $$I=\int \frac{\sin^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx$$ then $$I=\int \frac{-\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx+x$$ giving$$2I=\int \frac{\sin^{6}(x)-\cos^{6}(x)}{\sin^{6}(x) + \cos^{6}(x)} dx+x$$ This can be written as (using identities like $a^3-b^3$ and $a^3+b^3$) $$2I= \int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1-\sqrt3\sin(x)\cos(x)(1+\sqrt3\sin(x)\cos(x))}dx+x$$

$$ 2I=\frac{1}{2}\left(\int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1+\sqrt3\sin(x)\cos(x))} +\int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1-\sqrt3\sin(x)\cos(x))}\right)+x $$ Evaluating the integrals separately by using $u=1+\sqrt3\sin(x)\cos(x)$ for first one gives

$$\int\frac{(\sin^2(x)-\cos^2(x)(1-\sin^2(x)\cos^2(x))}{(1+\sqrt3\sin(x)\cos(x))}=\frac{1}{\sqrt3}\int\frac{(\sin(x)\cos(x)-1)(\sin(x)\cos(x)+1)}{u}du$$ Now use $\sin(x)\cos(x)=\frac{u-1}{\sqrt3}$ which will evaluate the integral as $\frac{u^2}{6\sqrt3}-\frac{2u}{3\sqrt3}-\frac{2\ln(u)}{3\sqrt3}$...Similar approach for other with $v=1-\sqrt3\sin(x)\cos(x)$

The final value is $$I=\frac{x}{2}-\frac{\sin(x)\cos(x)}{6}+\frac{\ln(1-\sqrt3\sin(x)\cos(x))}{6\sqrt3}-\frac{\ln(1+\sqrt3\sin(x)\cos(x))}{6\sqrt3}+C$$

  • $\begingroup$ I dont see how partial fraction will be easier now? $\endgroup$ – King Tut Mar 1 '18 at 11:26
  • $\begingroup$ I have tried the same approach on my notes but it became lengthy. Still liked it for alternate approach! $\endgroup$ – King Tut Mar 1 '18 at 11:37

An alternative would be to do a load of trig manipulation:

Starting with $$\frac{s^6}{s^6+c^6}=1-\frac{c^6}{s^6+c^6}$$ $$=1-\frac{\frac 18(1+\cos2x))^3}{ \frac 18(1+\cos2x))^3+\frac 18(1-\cos2x))^3}$$ = a few more lines of algebra = $$=\frac 12+\frac 12\cos 2x\left(\frac{3+\cos^2 2x}{1+3\cos^2 2x}\right)$$ $$=\frac 12+\frac 12\cos 2x\left(\frac{4-\sin^2 2x}{4-3\sin^2 2x}\right)$$

Now integrate, and for the second part us the substitution $$u=\frac{\sqrt{3}}{2}\sin 2x$$

and you end up with $$\frac 12 x+\frac 16\sin x\cos x+\frac{1}{3\sqrt{3}}\operatorname{artanh}(\sqrt{3}\sin x\cos x)+c$$

  • $\begingroup$ Thanks, for the insight. Some few algebra is by factorize the denominator is it? $\endgroup$ – Arief Anbiya Mar 2 '18 at 3:46
  • $\begingroup$ @Arief. You just have to multiply out the brackets and cancel terms $\endgroup$ – David Quinn Mar 2 '18 at 10:41

The partial fractions are not difficult to find if you use $z$ to denote $u^2$. This makes no difference as we are just using it to find partial fraction.

$$\frac{1}{(1+u^6)(1+u^2)} = \frac{1}{(1+z)^2(z^2-z+1)} \\= \frac{A}{1+z}+\frac{B}{(1+z)^2} + \frac{Cz+D}{z^2-z+1}$$

Finding $A$ is simple using heaviside cover rule, we get $B= \frac{1}{3}$. Now $A,C,D$ follow from three equations, which I leave as a task for you.

Upon finding the coefficients, again denote $z$ as $u^2$ and integrate.


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