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i would appreciate if you could help me with this problem. $$I=\int_{0}^{\pi}\frac{x \sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}dx$$ I am completely overwhelmed on how to proceed with this and i am stuck. so i would appreciate some tips to start

after reading some tips i have done this $$\int_{0}^{\pi}\frac{(\pi-x) \sin^{2018}(\pi-x)}{\cos^{2018}(\pi-x)+\sin^{2018}(\pi-x)}dx$$

after that i distributed it $$\int_{0}^{\pi}\frac{\pi\sin^{2018}(\pi-x)-x\sin^{2018}(\pi-x)}{\cos^{2018}(\pi-x)+\sin^{2018}(\pi-x)}=\int_{0}^{\pi}\frac{\pi\sin^{2018}(x)}{\cos^{2018}(x)+\sin^{2018}(x)}-I$$ But after that i'm lost again, i know i should be look for a way to make some terms cancel but that's as far as i got

please don't give me the direct solution, i would like to solve it by myself

Thanks in advance!

Ramiro genta
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  • without the $\pi$ in the numerator, the last integral in the last line (not the $-I$) evaluates to $\pi/2$, if i recall correctly – Benjamin Wang Mar 06 '21 at 20:04
  • Use https://www.google.com/amp/s/amp.doubtnut.com/question-answer/if-f-is-an-integrable-function-such-that-f2a-xfx-then-prove-that-int02afxdx2int0afxdx-1462410 and https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 – lab bhattacharjee Mar 06 '21 at 20:08
  • For the last unknown integral in the last line Use substitution pi/2 - x on interval 0 to \pi / 2. The other interval gives same answer – Benjamin Wang Mar 06 '21 at 20:11
  • I arrivied at the answer \frac{\pi}{2} is it correct? – Ramiro genta Mar 06 '21 at 20:22
  • First of all your need to replace $2018$ by $2n$ to simplify typing. $2n$ is used because the fact that $2018$ is even is crucial here. Your approach is correct and leads to $I=(\pi/2)\int_0^{\pi/2}\sin^{2n}x/(\sin^{2n}x+\cos^{2n}x),dx$. You can now use the standard way to replace upper limit by $\pi/2$ and after the answer should be obvious. – Paramanand Singh Mar 07 '21 at 02:17

1 Answers1

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HINT

Note that for even $n$ (as in your case) we have $$\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx=\int_0^{\pi} \frac{\cos^n x}{\sin^n{x}+\cos^n x}dx$$

As asked for, this is a hint :) If you need any more help please don't hesitate to ask.

This is how I'd continue. We know that $$\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx=\int_0^{\pi} \frac{\cos^n x}{\sin^n{x}+\cos^n x}dx$$ Adding, we find that $$2\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx=\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx+\int_0^{\pi} \frac{\cos^n x}{\sin^n{x}+\cos^n x}dx=\pi$$ So $$\int_0^{\pi} \frac{\sin^n x}{\sin^n{x}+\cos^n x}dx=\frac{\pi}{2}$$ and in particular $$\int_0^{\pi} \frac{\sin^{2018} x}{\sin^{2018}{x}+\cos^{2018}x}dx=\frac{\pi}{2}$$ We now know that $$I=\pi\int_0^{\pi} \frac{\sin^{2018} x}{\sin^{2018}{x}+\cos^{2018}x}dx-I=\pi\times\frac{\pi}{2}-I$$ Hence $2I=\pi\times\frac{\pi}{2}=\frac{\pi^2}{2}$ and $I=\frac{\pi^2}{4}$.

A-Level Student
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  • Why should this equality hold? Perhaps you were thinking that upper limit of integrals is $\pi/2$. – Paramanand Singh Mar 07 '21 at 02:18
  • @ParamanandSingh no, I'm quite certain this equality holds. Note that I specified that it's true for even $n$ only. I believe it holds because for even $n$, $\cos^nx$ and $\sin^n x$ have a period of $\pi$, and from $0$ to $\pi$ they are the same, just shifted by $\pi/2$, which makes no difference when it comes to the area bounded by them over a full period as they are always positive. Try it out on WA for different even values of $n$ to convince yourself :) – A-Level Student Mar 07 '21 at 09:35
  • @Ramirogenta I'm so glad I could help! Do you have any questions? – A-Level Student Mar 07 '21 at 10:02
  • Oh my bad!! You caught me there! Since the equality for $[0,\pi/2]$ is obvious and the integral for $[0,\pi]$ is twice of that over $[0,\pi/2]$ (for even $n$), your result holds. +1 – Paramanand Singh Mar 07 '21 at 10:07
  • @ParamanandSingh thank you, I also like your way of looking at it. – A-Level Student Mar 07 '21 at 10:08
  • Well if you could correct this it would be incredible, so i did this $ \int_0^{\pi} \frac{x\sin^n x}{\sin^n{x}+\cos^n x}dx+\int_0^{\pi} \frac{x\cos^n x}{\sin^n{x}+\cos^n x}dx$ and it gave me $$(\int_{0}^{\pi}x, dx) 1/2=\frac{\pi^2}{4}$$ is this the correct way? – Ramiro genta Mar 07 '21 at 10:09
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    @Ramirogenta: the answer is indeed $\pi^2/4$. – Paramanand Singh Mar 07 '21 at 10:11
  • @Ramirogenta I don't know how to justify your method to be honest although intuitively it seems quite sound; it's probably right. Perhaps Paramanand Singh can help? I'll update my answer to show how I'd do it. – A-Level Student Mar 07 '21 at 10:14
  • @ParamanandSingh please could you justify the OP's method? I don;t fully understand why it works. – A-Level Student Mar 07 '21 at 10:14
  • Well, I doubt if asker's approach is correct. It's just that somehow the answer came correct. – Paramanand Singh Mar 07 '21 at 10:19
  • @ParamanandSingh ok, thanks. Maybe I'll ask a question about it. – A-Level Student Mar 07 '21 at 10:23
  • I'm sorry, but how did you end up doing multiplying $\pi \times \frac{\pi}{2}$? isin't already know that $2I=\pi$ so why is the answer not $I=\frac{\pi}{2}$ ? – Ramiro genta Mar 07 '21 at 10:47
  • @Ramirogenta see my edit. If you're still unclear please tell me. – A-Level Student Mar 07 '21 at 10:51
  • Oh, ok thaks, now its clear, thanks for the help! – Ramiro genta Mar 07 '21 at 10:54
  • @Ramirogenta you are very welcome! :) – A-Level Student Mar 07 '21 at 10:56