1. ## Trigonometric substitution

Evaluate the indefinite integral.

I'm having trouble how to start. Do I factor out the x to make it ∫√x√16-x dx?

2. Originally Posted by emurphy
Evaluate the indefinite integral.

I'm having trouble how to start. Do I factor out the x to make it ∫√x√16-x dx?
Complete the square: $16x-x^2=64-64+16x-x^2=64-(x-8)^2$

So you have $\int\sqrt{64-(x-8)^2}\,dx$

Now let $x=8+8\sin t, dx=8\cos t\,dt$, so you get:

$\int\sqrt{64-64\sin^2t}\cdot8\cos t\,dt=64\int\cos^2t\,dt=64\int\frac{\cos(2t)+1}{2} \,dt$

Integrate and plug back in for $t$ using $t=\sin^{-1}\left(\frac{x-8}{8}\right)$.

3. I got this to be my answer

and the integral came out to be sin2t+x+c is this correct?

4. Originally Posted by emurphy
I got this to be my answer

and the integral came out to be sin2t+x+c is this correct?
You can check your answer by differentiating it. You can also use this website: Wolfram|Alpha

5. it didnt have it but could you check the if i did the integration for $dt=64\int\frac{\cos(2t)+1}{2}\,dt$ correctly?

6. Originally Posted by emurphy
it didnt have it but could you check the if i did the integration for $dt=64\int\frac{\cos(2t)+1}{2}\,dt$ correctly?
You can check it at the website whose link I gave you.