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I am trying to solve this integral by inverse trigonometric substitution. $$ \int_{0}^1{\sqrt{4kx-k^2x^2}dx} $$ where k is an arbitrary constant. I completed the square to get this:$$ \int_{0}^1\sqrt{2^2-(kx-2)^2}dx $$ Then, I substituted $kx-2=2\sin \theta$ into the integral to obtain: $$ \int_{-\pi/2}^{\sin^{-1}(\frac{k-2}{2})}2\cos{\theta}\frac{2}{k}\cos\theta d\theta \\ =\frac{2}{k}\int_{-\pi/2}^{\sin^{-1}(\frac{k-2}{2})}2\cos^2{\theta}+1-1d\theta \\ =\frac{1}{k}\int_{-\pi/2}^{\sin^{-1}(\frac{k-2}{2})}2\cos(2\theta)+2d\theta \\ =\frac{1}{k}(\sin(2\theta)+2\theta)|_{-\pi/2}^{\sin^{-1}(\frac{k-2}{2})} $$ However, when I tried to check by substituting $k=2$, I get$$ \int_{0}^1{\sqrt{4\cdot2\cdot x-2^2x^2}dx}=1.333333754 $$ but my answer returns$$ \frac{1}{k}(\sin(2\theta)+2\theta)|_{-\pi/2}^{-\sin^{-1}(\frac{k-2}{2})}=1.570796327$$ I am not sure where I have gone wrong with my substitution. I thought I merely left out a 2 multiple but that was not the case. Thank you very much for your help.

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    $\begingroup$ Why is there no $k$ in the boundaries? $\endgroup$
    – vitamin d
    Feb 22, 2021 at 13:47
  • $\begingroup$ Are you sure this is a correct integral? If $k$ is arbitrary, you could get something uncomfortable in the case k>4. Or k<0 $\endgroup$
    – Yuriy S
    Feb 22, 2021 at 13:50
  • $\begingroup$ I apologise for that, I have edited in the boundaries. The lower bound remained the same because $x=0$. There was no information provided for k in the question, I could only assume it to be properly behaving. $\endgroup$
    – Memiya
    Feb 22, 2021 at 13:57

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You want to substitute $kx-2=2\sin\theta$. For the lower boundary, $x=0$, we have $$-2=2\sin\theta\Rightarrow\theta=-\frac{\pi}{2}.$$ And for $x=1$, we have $$k-2=2\sin\theta\Rightarrow\theta=\arcsin\left(\frac{k}{2}-1\right).$$

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  • $\begingroup$ Yes, I did this, and when I checked k=2, I got the answer above of 1.57 instead of 1.33. I am not sure where I went wrong now. $\endgroup$
    – Memiya
    Feb 22, 2021 at 14:11
  • $\begingroup$ I tried to do that, but I ended up with this: $$ \frac{1}{k}\int_{-\pi/2}^{\arcsin(\frac{k}{2}-1)}{2\cos(2\theta)+2 d\theta}=1.5708 $$ Note: Sorry about that, I accidentally hit enter instead of shift+enter. This is when k=2. $\endgroup$
    – Memiya
    Feb 22, 2021 at 14:34
  • $\begingroup$ I can't see the formula. $\endgroup$
    – vitamin d
    Feb 22, 2021 at 14:35
  • $\begingroup$ Sorry for the ping, but I just needed to notify you that I updated my comment with the formula. $\endgroup$
    – Memiya
    Feb 22, 2021 at 15:22
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    $\begingroup$ I really am sorry for wasting your time. It seems that a calculator reset solved the issue. I got $\pi/2$ with the original now. Thank you very much for your time and help, I really appreciate it. $\endgroup$
    – Memiya
    Feb 22, 2021 at 15:32

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