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$$\int \frac{dx}{\sqrt{-x^2-12x+28}}$$

First we need to use completing the square $-(x^2+12x-28)=-(x+6)^2+64$

So we have $\int \frac{dx}{\sqrt{-(x+6)^2+64}}$ I know that it is a general form of $\arcsin(\frac{x+6}{8})$ but how can I solve it using substitution?

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  • $\begingroup$ try $\sin \theta = \frac{x+6}{8}$ $\endgroup$ – vnd Jan 4 '16 at 18:45
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Put $y=\frac{x+6}{8}$ then the integral is equivalent to $$ \int \frac{\mathrm{d}y}{\sqrt{1-y^2}}. $$ From here you can conclude..

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Its not really necessary to go for substitution. Use basic differentiation of inverse trig functions.

$\begin{align} \int\frac{dx}{\sqrt{a^2-x^2}}&= \sin^{-1}(\frac{x}{a})+C\\ &=\int\frac{dx}{\sqrt{64-(x+6)^2}}\\ &=\sin^{-1}\frac{(x+6)}{8}+C \end{align} $

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  • $\begingroup$ You forgot square roots in your integrals. $\endgroup$ – mickep Jan 6 '16 at 10:02
  • $\begingroup$ '@mickep, Thanks.The answer does not change much though,but thanks again.' $\endgroup$ – Tosh Jan 7 '16 at 7:04

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