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I am facing trouble to prove: $$\displaystyle\int\frac{\mathrm d x}{\sqrt{a+bx+cx^2}}= \frac{1}{\sqrt{-c}}\arccos\left(-\frac{b+2cx}{\sqrt{-\Delta}} \right) \tag{1}\\ \text{when } c<0, \; \Delta=4ac-b^2<0 $$ It's used here: https://youtu.be/O4SIw6cYSow?t=577

But, so far, I have come up with this form: $$\displaystyle\int\frac{\mathrm d x}{\sqrt{a+bx+cx^2}}= \frac{-1}{\sqrt{-c}}\arcsin\left(\frac{b+2cx}{\sqrt{-\Delta}} \right)= \frac{1}{\sqrt{-c}}\arccos\left(\frac{b+2cx}{\sqrt{-\Delta}} \right) \tag{2}$$ As you can see, the necessary minus sign isn't appearing inside the inverse cosine function!

By the way, the form with arcsine function is also mentioned in section 2.261 of the book "Table of Integrals, Series, and Products". (https://www.amazon.com/Table-Integrals-Products-Daniel-Zwillinger/dp/0123849330)

TIA

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  • $\begingroup$ Maybe you or the YouTube video has made a silly mistake. Mind sharing your steps? $\endgroup$ Nov 14, 2019 at 10:46
  • $\begingroup$ @Certainlynotadog I don't think the Youtube video had done any mistake. The formula is used there to derive an equation that was correct. Meanwhile, I have got the exact arcsine form as it is mentioned in the book "Table of Integrals, Series, and Products". But the arccosine form isn't matching with the Youtube video's. I will show my steps later. Can you please check which one seems correct to you? $\endgroup$
    – raf
    Nov 14, 2019 at 11:01
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    $\begingroup$ Where is the integration constant? $\endgroup$
    – Arthur
    Nov 14, 2019 at 11:09
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    $\begingroup$ Hint : What is the relationship between $\arccos(t)$ and $\arccos(-t),$ and how is it connected to your omission of an integration constant ? $\endgroup$
    – Lucian
    Nov 14, 2019 at 11:25
  • $\begingroup$ $\arccos{(-t)}=\pi - \arccos{(t)}$. I just want to derive the equation (1) keeping $\theta_o$ as the integration constant. Is it wrong? @Lucian $\endgroup$
    – raf
    Nov 14, 2019 at 12:13

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