2

My first guess is it can't, since when I substitute the boundaries, I end up with $\cot2\pi$ and $\cot0$. Nevertheless I tried substituting pretending it is indefinite integral, but I can't get anything meaningful.

Michael Hardy
  • 1
Stefan Stipanovic
  • 267
  • 2
    $\cot(2\pi)$ and $\cot0$ are both $\infty$ (neither $+\infty$ nor $-\infty$, but the $\infty$ that's at both ends of the line). However, some symmetries show that $$ \int_0^{2\pi} \frac{dx}{\sin^x + \cos^6x},dx = 4\int_0^{\pi/2} \frac{dx}{\sin^6 x + \cos^6x},dx. $$ On that interval, $\cot x$ goes from $\infty$ down to $0$. $\qquad$ – Michael Hardy Jun 19 '16 at 18:55
  • @MichaelHardy How do you get to second interval? Also, could you write how would you go about solving this integral with $\cot x=u$ as substitution? Thanks! – Stefan Stipanovic Jun 19 '16 at 19:02
  • any rational function of $\sin x, \cos x$ can be changed to a rational function with https://en.wikipedia.org/wiki/Tangent_half-angle_substitution – Will Jagy Jun 19 '16 at 19:08
  • The function being integrated has period $\pi$; therefore its integral over $[0,\pi]$ is the same as its integral over $[\pi,2\pi]$. Next, we have $\cos\left(\frac \pi 2 - x\right) = \sin x$ and $\sin\left(\frac \pi 2 - x\right) = \cos x$, so the substitution $v=\frac\pi2-x$ transforms $$ \int_{\pi/2}^\pi $$ into $$ \int_0^{\pi/2} $$ with sine and cosine interchanged. But interchanging $\sin$ and $\cos$ in $\sin^6x+\cos^6x$ gives you $\cos^6x+\sin^6x$, which is the same function. Therefore $$ \int_{\pi/2}^\pi = \int_0^{\pi/2}.$$ – Michael Hardy Jun 19 '16 at 19:09

4 Answers4

5

Using $$ \int_0^\infty\frac{x^n}{1+x^m}dx= \frac{\pi \csc \left(\frac{\pi (n+1)}{m}\right)}{m} $$ (see Prove $\int_0^{\infty}\! \frac{\mathbb{d}x}{1+x^n}=\frac{\pi}{n \sin\frac{\pi}{n}}$ using real analysis techniques only) it is easy to see \begin{eqnarray} \int_0^{2\pi} \frac{dx}{\sin^6x+\cos^6x}&=&4\int_0^{\frac{\pi}{2}} \frac{dx}{\sin^6x+\cos^6x}\\ &=&4\int_0^{\frac{\pi}{2}} \frac{\csc^6x\,dx}{1+\cot^6x}\\ &=&4\int_0^{\frac{\pi}{2}} \frac{\csc^4x}{1+\cot^6x}\csc^2x\,dx\\ &=&4\int_0^{\frac{\pi}{2}} \frac{(1+\cot^2x)^2}{1+\cot^6x}\csc^2x\,dx\\ &=&4\int_0^\infty\frac{(1+u^2)^2}{1+u^6}\,du\\ &=&\cdots \end{eqnarray}

Community
  • 1
xpaul
  • 44,000
3

We have $$\int_{0}^{2\pi}\frac{1}{\cos^{6}\left(x\right)+\sin^{6}\left(x\right)}dx=4\int_{0}^{\pi/2}\frac{1}{\cos^{6}\left(x\right)+\sin^{6}\left(x\right)}\cdot\frac{\cos^{4}\left(x\right)}{\cos^{4}\left(x\right)}dx\tag{1}$$ $$=4\int_{0}^{\pi/2}\frac{\left(\tan^{2}\left(x\right)+1\right)\sec^{2}\left(x\right)}{\tan^{4}\left(x\right)-\tan^{2}\left(x\right)+1}dx\overset{\tan\left(x\right)=u}{=}4\int_{0}^{\infty}\frac{u^{2}+1}{u^{4}-u^{2}+1}du $$ and now observe that $$\int_{0}^{\infty}\frac{u^{2}+1}{u^{4}-u^{2}+1}du=\int_{0}^{\infty}\frac{u^{2}+1}{\left(u^{2}-1\right)^{2}+u^{2}}du $$ $$=\int_{0}^{\infty}\frac{\frac{u^{2}+1}{u^{2}}}{\left(\frac{u^{2}-1}{u}\right)^{2}+1}du\overset{\frac{u^{2}-1}{u}=v}{=}\int_{-\infty}^{\infty}\frac{1}{1+v^{2}}dv=\pi $$ hence

$$\int_{0}^{2\pi}\frac{1}{\cos^{6}\left(x\right)+\sin^{6}\left(x\right)}dx=4\pi.$$

Note: In $(1)$ we can multiply by $\frac{\sin^{4}\left(x\right)}{\sin^{4}\left(x\right)}$ instead of $\frac{\cos^{4}\left(x\right)}{\cos^{4}\left(x\right)}$ and get $$4\int_{0}^{\pi/2}\frac{\left(\cot^{2}\left(x\right)+1\right)\csc^{2}\left(x\right)}{\cot^{4}\left(x\right)-\cot^{2}\left(x\right)+1}dx$$ and now we can take $\cot(x)=u$. The rest is the same.

Marco Cantarini
  • 33,062
  • 2
  • 47
  • 93
  • Thanks for your answer. I knew of that way to solve this one. But the question asks can it be solved using $\cot x=u$. Otherwise I wouldn't have posted this – Stefan Stipanovic Jun 19 '16 at 21:16
  • I guess it would be the same if multiplied by $\csc^4x$ – Stefan Stipanovic Jun 19 '16 at 21:17
  • @StefanStipanovic Sorry I read too fast. As you said is the same if you multiply by $\frac{\sin^{4}(x)}{\sin^{4}(x)}$. Instead of $\tan$ you will get $\cot$. Then make the substitution $\cot(x)=u$. – Marco Cantarini Jun 19 '16 at 21:32
1

Hint $$\sin^6x+\cos^6x=1-\frac{3}{4}\sin^2 2x$$

Behrouz Maleki
  • 11,182
1

If we consider the problem of the antiderivative $$I=\int \frac{dx}{\sin^6(x)+\cos^6(x)}$$ and use $$\cot(x)=u\implies x=\cot ^{-1}(u)\implies dx=-\frac{du}{u^2+1}$$ after some minor simplifications $$I=-\int\frac{u^2+1}{u^4-u^2+1}\,du=\tan ^{-1}\left(\frac{u}{u^2-1}\right)$$ (just as Marco Cantarini answered).

For the definite integral, use Michael Hardy's comment.

Claude Leibovici
  • 260,315