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$$\int_{0}^{\frac{\pi}{2}} \frac{\cos^5(x)}{\sin^5(x) + \cos^5(x)} \,dx$$

I tried by dividing the terms in both the numerator and denominator by $\cos^5x$ but still cant find my way.

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  • $\begingroup$ I am not sure what integral you are looking at. $\endgroup$
    – user357980
    Jun 14, 2017 at 2:57
  • $\begingroup$ Were you given limits by any chance? $\endgroup$
    – mrnovice
    Jun 14, 2017 at 3:16
  • $\begingroup$ yeah from zero to pi over two $\endgroup$
    – Acade
    Jun 14, 2017 at 3:19
  • $\begingroup$ @Acade that is extremely important information which greatly simplifies the question $\endgroup$
    – mrnovice
    Jun 14, 2017 at 3:19
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    $\begingroup$ I'm downvoting to discourage this sort of omission of important information from questions (until asked explicitly). It causes those who answer to do a lot of unnecessary work, and isn't fair. $\endgroup$
    – Deepak
    Jun 14, 2017 at 3:34

2 Answers 2

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$$I =\int_{0}^{\frac{\pi}{2}} \frac{\cos^5x}{\sin^5x + \cos^5x} dx\tag{1}$$

Let $u =\frac{\pi}{2}-x$

$$I = -\int_{\frac{\pi}{2}}^{0} \frac{\sin^{5}u}{\sin^5u+\cos^5u}du\tag{2} =\int_{0}^{\frac{\pi}{2}} \frac{\sin^5x}{\sin^5x + \cos^5x} dx$$

Then $$(1)+(2)\implies 2I=\int_{0}^{\frac{\pi}{2}}\frac{\sin^5x+\cos^5x}{\sin^5x+\cos^5x}dx=\int_{0}^{\frac{\pi}{2}}1dx=\frac{\pi}{2}\implies I=\frac{\pi}{4}$$

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  • $\begingroup$ But it's another problem. Your integral even I can calculate! $\endgroup$ Jun 14, 2017 at 3:24
  • $\begingroup$ @MichaelRozenberg Yes it is, but I suspected the OP had missed out information in the question and it turns out he did. $\endgroup$
    – mrnovice
    Jun 14, 2017 at 3:25
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    $\begingroup$ It's not fair! :D :-D $\endgroup$ Jun 14, 2017 at 3:32
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The problem was changed and my following work is not necessary.

Let $\tan{x}=t$.

Hence, $dt=\frac{1}{\cos^2x}dx=(1+t^2)dx$.

Thus, $$\int\frac{\cos^5x}{\sin^5x+\cos^5x}dx=\int\frac{1}{1+\tan^5x}dx=\int\frac{1}{(1+t^2)(1+t^5)}dt.$$

Let $t+\frac{1}{t}=u$.

Hence, $$1+t^5=(1+t)(1-t+t^2-t^3+t^4)=t^2(1+t)\left(t^2+\frac{1}{t^2}-t-\frac{1}{t}+1\right)=$$ $$=t^2(1+t)(u^2-2-u-1)=t^2(1+t)(u^2-u-1)=t^2(1+t)\left(u-\frac{1-\sqrt5}{2}\right)\left(u-\frac{1+\sqrt5}{2}\right)=$$ $$=(1+t)\left(t^2-\frac{1-\sqrt5}{2}t+1\right)\left(t^2-\frac{1+\sqrt5}{2}t+1\right).$$ The rest is smooth:

Let $$\frac{1}{(1+t)(1+t^2)(\left(t^2-\frac{1-\sqrt5}{2}t+1\right)\left(t^2-\frac{1+\sqrt5}{2}t+1\right)}=$$ $$=\frac{A}{1+t}+\frac{Bt+C}{1+t^2}+\frac{Dt+E}{t^2-\frac{1-\sqrt5}{2}t+1}+\frac{Ft+G}{t^2-\frac{1+\sqrt5}{2}t+1}$$ and solve the system.

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  • $\begingroup$ the resulting integral is still difficult to get through for me please can you help me with a solution? $\endgroup$
    – Acade
    Jun 14, 2017 at 3:11
  • $\begingroup$ @Acade I fixed my post, See now, $\endgroup$ Jun 14, 2017 at 3:16
  • $\begingroup$ @Acade Did you copy the question with all the given information? Are you sure you weren't given limits for the integral? $\endgroup$
    – mrnovice
    Jun 14, 2017 at 3:18
  • $\begingroup$ thanks alot i am trying to see the system given be solve $\endgroup$
    – Acade
    Jun 14, 2017 at 3:20
  • $\begingroup$ limits from zero to pi over two $\endgroup$
    – Acade
    Jun 14, 2017 at 3:20

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