`int (x+5)/sqrt(9-(x-3)^2) dx` Find the indefinite integral

Friday, December 9, 2011

$\displaystyle\int\frac{{{x}+{5}}}{\sqrt{{{9}-{{\left({x}-{3}\right)}}^{{2}}}}}{\left.{d}{x}\right.}$ Find the indefinite integral

We have to evaluate the integral $\int \frac{x+5}{\sqrt{9-(x-3)^2}}dx$

Let $\displaystyle{x}-{3}={u}$

So, $\displaystyle{\left.{d}{x}\right.}={d}{u}$

Hence we can write,

$\int \frac{x+5}{\sqrt{9-(x-3)^2}}dx=\int \frac{u+8}{\sqrt{9-u^2}}du$

$=\int \frac{u}{\sqrt{9-u^2}}du+\int \frac{8}{\sqrt{9-u^2}}dx$

Now we will first evaluate the integral: $\int \frac{udu}{\sqrt{9-u^2}}$

Let $\displaystyle{9}-{{u}}^{{2}}={t}$

So, $\displaystyle-{2}{u}{d}{u}={\left.{d}{t}\right.}$

Hence we have,

$\int \frac{udu}{\sqrt{9-u^2}}=\int \frac{-dt}{2\sqrt{t}}$

$=-\sqrt{t}$

$=-\sqrt{9-u^2}$

Now we will evaluate the integral $\int \frac{8}{\sqrt{9-u^2}}du$

$\int \frac{8}{\sqrt{9-u^2}}du=\frac{8du}{3\sqrt{1-(\frac{u}{3})^2}}$

$=8sin^{-1}(\frac{u}{3})$ since we have the identity

$\frac{d}{dx}(sin^{-1}(\frac{u}{a}))=\frac{1}{a}.\frac{1}{\sqrt{1-(\frac{u}{a})^2}}$

So finally we have the result as:

$\int \frac{x+5}{\sqrt{9-(x-3)^2}}dx=-\sqrt{9-u^2}+8sin^{-1}(\frac{u}{3})+C$

$=-\sqrt{9-(x-3)^2}+8sin^{-1}(\frac{x-3}{3})+C$

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