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I'm sorry if this is a simple question. if anyone can correct me, I'd greatly appreciate it. So first i used substitution $x = sin\theta$ so we have

$$\int _0^{\frac{\pi }{2}}\frac{\ 1}{\sqrt{\ 1-\sin ^2\theta }}\sin ^{-1}\left(2\sin \theta \cos \theta \ \right)\cos \theta d\theta$$

$$\int _0^{\frac{\pi }{2}}\ \sin ^{-1}\left(\sin 2\theta \right)d\theta $$ cancelling out $\sin^{-1}(\sin 2\theta)$ we are only left with $2\theta$ so aren't we supposed to get $$\int _0^{\frac{\pi }{2}}\ 2\theta d\theta $$ which leaves us with $2\left[\frac{\ \theta ^2}{2}\right]$ for bound from 0 till $\frac{\pi }{2}$ so I get $\frac{\pi ^2}{4}$ but my answer is wrong according to my text book it says it should be $\frac{\pi ^2}{8}$ but how? where did i go wrong?

I even shaded the area we are interested in finding area of enter image description here

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    $\begingroup$ In $\int _0^{\frac{\pi }{2}}\ \sin ^{-1}\left(\sin 2\theta \right)d\theta$ perform the change of variable $x=2\theta$ and look at the interval of definition of $arcsin$ function. $\endgroup$
    – FDP
    Oct 5 '20 at 8:32
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    $\begingroup$ Note that $\arcsin(\sin x)=x$ hold true for all $0\leq x\leq \frac{\pi}{2}$ however, $ 0\leq 2x \leq \pi$ so the $\arcsin(\sin2x)\neq 2x$ $\endgroup$
    – Naren
    Oct 5 '20 at 8:33
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    $\begingroup$ Ohhh Noted @Naren Thank you. $\endgroup$
    – RiRi
    Oct 5 '20 at 8:41
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Your error is in your attempt to simplify $\sin^{-1}\bigl(\sin(2\theta)\bigr)$ to $2\theta$.

For $0\le\theta\le{\large{\frac{\pi}{2}}}$, we get $$ \sin^{-1}\bigl(\sin(2\theta)\bigr) = \begin{cases} 2\theta&\text{if}\;0\le\theta\le\frac{\pi}{4}\\[4pt] \pi-2\theta&\text{if}\;\frac{\pi}{4}\le\theta\le\frac{\pi}{2}\\ \end{cases} $$ so you need to sum two integrals, one for each of the above cases, yielding $$ \int_0^{\large{\frac{\pi}{4}}} (2\theta)\,d\theta + \int_{\large{\frac{\pi}{4}}}^{\large{\frac{\pi}{2}}} (\pi-2\theta)\,d\theta = \frac{\pi^2}{16} + \frac{\pi^2}{16} = \frac{\pi^2}{8} $$

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In $\displaystyle \int _0^{\frac{\pi }{2}}\ \sin ^{-1}\left(\sin 2\theta \right)d\theta$ perform the change of variable $x=2\theta$ and look at the interval of definition of $\displaystyle \arcsin$ function.

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If you plot the integrand $$f(x) = \frac{\sin^{-1} 2x \sqrt{1-x^2}}{\sqrt{1-x^2}}$$ on $x \in [0,1]$, you will discover that there is a cusp around $x = 0.7$. Here is a picture:

enter image description here

What's going on? Well, on $[0,1]$, the function $\sqrt{1-x^2}$ is smooth, so that isn't the issue. This suggests that there is something going on with the inverse sine. Specifically, how do the plots of $$g(\theta) = \sin^{-1} (\sin 2\theta), \quad h(\theta) = 2\theta$$ compare? If $\theta = \pi/4$, then $\sin 2\theta = 1$, the local maximum. So when we take the inverse sine, we don't get $2\theta$ back; we get an angle in $[-\pi/4, \pi/4]$.

So to handle this issue, we have to be a bit more careful with the substitution. On $\theta \in [0,\pi/4]$, which corresponds to $x \in [0, 1/\sqrt{2}]$ (because the substitution is $x = \sin \theta$, hence $\theta = \pi/4$ means $x = 1/\sqrt{2}$), the integrand does indeed simplify to $$\int_{x=0}^{1/\sqrt{2}} f(x) \, dx = \int_{\theta = 0}^{\pi/4} 2\theta \, d\theta.$$ But on $x \in [1/\sqrt{2}, 1]$, we have instead $$\int_{x=1/\sqrt{2}}^1 f(x) \, dx = \int_{\theta = \pi/4}^{\pi/2} 2(\pi/2- \theta) \, d\theta.$$ These two pieces are equal; their sum is $\pi^2/8$ as claimed.

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Unfortunately the arcsin is a function and will only output values that are less than or equal to $\frac{\pi}{2}$. In the case of $\sin^{-1}(\sin(2\theta))$ for $0 \le \theta \le \frac{\pi}{2}$ the cancellation that you made of $\sin^{-1}(\sin(2\theta))=2\theta$ is invalid. As you can see on the graph: sin cancellation For this you must integrate from $0 \le \theta \le \frac{\pi}{4}$ and multiply the answer by two.

Not the most rigours explanation but hope this helps out a bit.

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$$I=\int_{0}^{\pi/2} \frac{dx}{\sqrt{1-x^2}} \sin^{-1} 2x\sqrt{1-x^2}$$ Let $x=\sin t, t=2u$ then $$I=\frac{1}{2}\int_{0}^{\pi} \sin^{-1}\sin u du =\int_{0}^{\pi/2} \frac{u}{2}+\int_{\pi/2}^{\pi} \frac{\pi-u}{2}du =\frac{\pi^2}{8}.$$

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