# Math Help - trig substitution problem

1. ## trig substitution problem

x²/sqrt(9-x²)

u = x
a = 3

i used.... x=asinx

i end up substituting this into the problem....
x = theta

(3cosx)²/3cosx

i then reduce it to

3cosx and integrate it to get....
3sinx + c

now i use my triangle and solve in terms of x and now i get as a final awnser...

x + c

this doesnt seem right, can anyone verify this for me!

2. It can't be right. The derivative of $x + C$ is not $\dfrac{x^2}{\sqrt{9-x^2}}$.

After performing the trigonometric substitution, the numerator should be $\left ( 3\sin{\theta} \right ) ^2$, not $\left ( 3\cos{\theta}{ \right ) ^2$.

3. ## ok

(9sinx² / 3cosx) dx

i integrated this properly i think and worked it to 9/3(ln|secx+tanx|-sinx) + C can anyone verify this for me?

4. $\displaystyle \int \frac{x^2}{\sqrt{9-x^2}} \, dx$

$x = 3\sin{t}$

$dx = 3\cos{t} \, dt$

$\displaystyle \int \frac{9\sin^2{t}}{\sqrt{9 - 9\sin^2{t}}} \cdot 3\cos{t} \, dt$

$\displaystyle \int \frac{9\sin^2{t}}{3\sqrt{1 - \sin^2{t}}} \cdot 3\cos{t} \, dt$

$\displaystyle \int 9\sin^2{t} \, dt$

$\displaystyle \frac{9}{2} \int 1 - \cos(2t) \, dt$

$\displaystyle \frac{9}{2} \left[t - \frac{\sin(2t)}{2}\right] + C$

$\displaystyle \frac{9}{4} \left[2t - \sin(2t)\right] + C$

$\displaystyle t = \arcsin\left(\frac{x}{3}\right)$

$\displaystyle \sin(2t) = 2\sin{t}\cos{t} = \frac{2x}{3} \cdot \sqrt{1 - \left(\frac{x}{3}\right)^2} = \frac{2x}{9} \sqrt{9 - x^2}$

so, antiderivative is ...

$\displaystyle \frac{9}{2}\left[\arcsin\left(\frac{x}{3}\right) - \frac{x \sqrt{9-x^2}}{9}\right] + C$