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Here is a question $\displaystyle \int \dfrac {\sin^\sqrt{11}x}{\sin^\sqrt{11}x + \cos^\sqrt{11}x}dx$

It seems very difficult for me to calculate this form of the integral, but intuitively I observe that $\sin$ and $\cos$ are continous everywhere and obviously their power is absurd so I can specify integral as $\displaystyle \int _0 ^ {\pi/2} \dfrac {\sin^\sqrt{11}x}{\sin^\sqrt{11}x + \cos^\sqrt{11}x}dx$ since they are continous and positive on first quadrant.

Okay, what I did is following, I used the transformation $\sin x = \cos (\frac{\pi}{2} -x)$ but it is not working.

Thanks in advance.

Sine of the Time
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Fuat Ray
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  • The answer is $\pi/4$, this has been asked before – Sine of the Time Dec 24 '23 at 11:05
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    A more general result is $\int_0^{\pi/2}\frac1{1+\tan^k (x)}dx=\pi/4$ for $k\in \Bbb R$ – Sine of the Time Dec 24 '23 at 11:07
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    $\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}dx=\int_0^{\pi/2}\frac{\cos x}{\cos x +\sin x}=\frac{1}{2}\left(\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}dx+\int_0^{\pi/2}\frac{\cos x}{\cos x +\sin x}dx\right)=\frac12\int_0^{\pi/2}dx=\pi/4$. First step is motivated by the sub $x\mapsto \pi/2-x$. Note that $\sqrt{11}$ is a distractor – Sine of the Time Dec 24 '23 at 11:16
  • I disagree with your interpretation of the question presented at the start of your posting. The evaluation of $~\displaystyle \int f(x) ~dx~~$ is different from the evaluation of $~\displaystyle \int_a^b f(x) ~dx.~$ With the first question, you are attempting to find the anti-derivative $~F(x),~$ while the second question allows you to bypass searching for the anti-derivative and simply try to find the numerical evaluation of the integral using alternate analysis. – user2661923 Dec 24 '23 at 15:24
  • My question is not duplicate because my question was can we solve for it without boundaries? – Fuat Ray Dec 27 '23 at 18:06

1 Answers1

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The $\sqrt 11$ is arbitrary. Let us prove more generally that

$$\ I(n) = \int_0^{\frac{\pi}{2}} \frac{\sin^nx dx}{\sin^nx+\cos^nx} = \frac{\pi}{4} = \int_0^{\frac{\pi}{2}} \frac{\cos^nx dx}{\sin^nx+\cos^nx}$$

If we substitute $\ t = 0 + \frac{\pi}{2} - x$, then the integral becomes

$$\ I(n) = \int_{\frac{\pi}{2}}^0 \frac{-cos^nt dt}{sin^nt+cos^nt}$$

Now, we can swap the limits and multiply $-1$ to the integrand. Thus, we get

$$\ I(n) = \int_0^{\frac{\pi}{2}} \frac{cos^nt dt}{cos^nt+sin^nt}$$

Which is the same as if we were replacing $t$ by $x$ and rewriting:

$$\ I(n) = \int_0^{\frac{\pi}{2}} \frac{cos^nx dx}{cos^nx+sin^nx}$$

Now, add the two integrals. We get:

$$\ 2I(n) = \int_0^{\frac{\pi}{2}} \frac{\sin^nx dx}{\sin^nx+\cos^nx} + \int_0^{\frac{\pi}{2}} \frac{cos^nx dx}{cos^nx+sin^nx}$$

$$\ 2I(n) = \int_0^{\frac{\pi}{2}} 1 dx = \frac{\pi}{2}$$

This means that

$$\ I(n) = \frac{\pi}{4}$$

In fact, more generally, we can say that:

$$\ I = \int_a^b f(t) dt = \int_a^b f(a+b-t) dt$$

This is a well-known result used to solve symmetric integrals.