# How do you integrate int sqrt(5+4x-x^2) by trigonometric substitution?

Nov 25, 2016

The answer is $= \frac{9}{2} \arcsin \left(\frac{x - 2}{3}\right) + \frac{x - 2}{2} \sqrt{9 - {\left(x - 2\right)}^{2}} + C$

#### Explanation:

Let's complete the square

$5 + 4 x - {x}^{2} = 9 - {\left(2 - x\right)}^{2}$

Let $x - 2 = 3 \sin u$

$\mathrm{dx} = 3 \cos u \mathrm{du}$

$9 - {\left(2 - x\right)}^{2} = 9 - 9 {\sin}^{2} u = 9 {\cos}^{2} u$

$\int \sqrt{5 + 4 x - {x}^{2}} \mathrm{dx} = \int 3 \cos u \cdot 3 \cos u \mathrm{du}$

$= 9 \int {\cos}^{2} u \mathrm{du}$

$\cos 2 u = 2 {\cos}^{2} u - 1$

${\cos}^{2} u = \frac{1 + \cos 2 u}{2}$

$9 \int {\cos}^{2} u \mathrm{du} = \frac{9}{2} \int \left(1 + \cos 2 u\right) \mathrm{du}$

$= \frac{9}{2} \left(u + \frac{\sin 2 u}{2}\right) = \frac{9}{2} \left(u + \sin u \cos u\right)$

$\sin u = \frac{x - 2}{3}$

$u = \arcsin \left(\frac{x - 2}{3}\right)$

$\cos u = \frac{1}{3} \sqrt{9 - {\left(x - 2\right)}^{2}}$

$\frac{9}{2} \left(u + \sin u \cos u\right) = \frac{9}{2} \left(\arcsin \left(\frac{x - 2}{3}\right) + \frac{x - 2}{9} \sqrt{9 - {\left(x - 2\right)}^{2}}\right)$

$\int \sqrt{5 + 4 x - {x}^{2}} \mathrm{dx} = \frac{9}{2} \arcsin \left(\frac{x - 2}{3}\right) + \frac{x - 2}{2} \sqrt{9 - {\left(x - 2\right)}^{2}} + C$