Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \frac{1}{\sqrt{16-6 x-x^{2}}} d x$

Solution:

let I $=\int \frac{1}{\sqrt{16-6 x-x^{2}}} d x$

$=\int \frac{1}{\sqrt{-\left[x^{2}+6 x-16\right]}} d x$

$=\int \frac{1}{\sqrt{-\left[x^{2}+2 x(3)+(3)^{2}-(3)^{2}-16\right]}} d x$

$=\int \frac{1}{\sqrt{-\left[(x-3)^{2}-25\right]}} d x$

$=\int \frac{1}{\sqrt{25-(x+3)^{2}}} d x$

$\operatorname{let}(x+3)=t$

$\mathrm{d} \mathrm{x}=\mathrm{dt}$

$I=\int \frac{1}{\sqrt{5^{2}-t^{2}}} d t$

$=\sin ^{-1}\left(\frac{t}{5}\right)+c$

$I=\sin ^{-1}\left(\frac{x+3}{5}\right)+c$

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