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I have solved the problem and obtained the value of the integral $2\pi/3-\sqrt{3}/2$.

In the actual solution of this problem I have seen that $x$ was substituted by $2\sin z$ and upper limit $1$ was substituted by $\pi/6$ and lower limit $-1$ by $-\pi/6$.

Now the question is why was the upper limit substituted by $\pi/6$ instead of $5\pi/6$ or anything equivalent to $\arcsin\tfrac12$?

Kenta S
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MSKB
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  • You can actually use any such value for the integral bounds. The result comes out the same. Show that $\int_{\pi/6}^{5\pi/6} f(2\sin x)\cos x,dx =0$ for any $f(x).$ – Thomas Andrews Mar 09 '21 at 19:30

4 Answers4

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Actually,$$\sin(\pi-y)=\sin y\implies\sin\tfrac{5\pi}{6}=\sin\tfrac{\pi}{6}\ne-\sin\tfrac{\pi}{6}.$$We need a $z$ satisfying $\sin z=-\sin\tfrac{\pi}{6}$. Since the sine function is odd, $z=-\tfrac{\pi}{6}$ will do.

Alternative argument: since the original integrand is even,$$\begin{align}\int_{-1}^1x^2\sqrt{4-x^2}dx&=2\int_0^1x^2\sqrt{4-x^2}dx\\&=2\int_0^{\pi/6}16\sin^2z\cos^2zdz\\&=\int_{-\pi/6}^{\pi/6}16\sin^2z\cos^2zdz,\end{align}$$as the new integrand is even too.

J.G.
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When you think about the image of the interval $x \in [-1,1]$ under the mapping $$x = 2 \sin z,$$ we don't just want to find values of $z$ corresponding to the endpoints of the interval. We need to consider how the mapping affects the entire interval. If you plot $2 \sin z$, you can see that there are infinitely many possible ways that the interval $x \in [-1,1]$ can be mapped to some $z$-interval; for instance, you could choose $z \in [5 \pi/6, 7\pi/6]$, but this also reverses the order of integration; or you could choose $z \in [11 \pi/6, 13 \pi/6]$, etc. But you cannot choose, for instance, $z \in [-\pi/6, 5\pi/6]$, because this interval contains points that are not mapped to $x \in [-1,1]$, e.g., $z = \pi/2$ maps to $x = 2$.

heropup
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  • The value of z remains the same if we choose the interval [5pi/6,-pi/6],but this interval will affect the value of integral. – MSKB Mar 09 '21 at 19:02
  • So does the value of this integral depend on the interval chosen by the one who is solving?Since for infinite number of intervals the mappimg of Z remains constant yet the value of integral changes – MSKB Mar 09 '21 at 19:10
  • @MohammadSakibShahriar I have already explained this in my answer. If you do not understand you need to be more specific about what you do not understand. – heropup Mar 09 '21 at 19:12
  • I apologize for disturbing you.....thank you very much for your kindness.......Now I really understood the whole thing.......Thanks again – MSKB Mar 09 '21 at 19:39
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In general, if $f(x)$ is any function defined on $[-1,1]$ and $\sin \theta_1=\sin\theta_2$ then $$\int_{\theta_1}^{\theta_2} f(\sin \theta)\cos(\theta)\,d\theta=0$$ So it doesn’t matter which $\arcsin 1/2$ you choose.

If $f$ was only defined on $[-1/2,1/2],$ you’d have to pick a careful pair of integral bounds. But any pair such that $f(\sin\theta)$ is defined between the pairs would work.

Or you could cheat and define $f(x)=0$ outside $[-1/2,1/2].$

Thomas Andrews
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It must be $2 \sin{z} = \pm1 \Rightarrow z = \arcsin{\pm\frac{1}{2}} = \pm\frac{\pi}{6}$ That's all.

Lizzi
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