Evaluate the indefinite integral.

http://hosted.webwork.rochester.edu/...b0a24d46a1.png

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

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- Oct 19th 2009, 07:38 PMemurphyTrigonometric substitutionEvaluate the indefinite integral.

http://hosted.webwork.rochester.edu/...b0a24d46a1.png

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

- Oct 19th 2009, 07:46 PMredsoxfan325
Complete the square: $\displaystyle 16x-x^2=64-64+16x-x^2=64-(x-8)^2$

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

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

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

Integrate and plug back in for $\displaystyle t$ using $\displaystyle t=\sin^{-1}\left(\frac{x-8}{8}\right)$. - Oct 19th 2009, 07:57 PMemurphy
I got this to be my answer

http://hosted.webwork.rochester.edu/...466a0d8de1.png

and the integral came out to be sin2t+x+c is this correct? - Oct 19th 2009, 09:15 PMmr fantastic
You can check your answer by differentiating it. You can also use this website: Wolfram|Alpha

- Oct 19th 2009, 09:51 PMemurphy
it didnt have it but could you check the if i did the integration for $\displaystyle dt=64\int\frac{\cos(2t)+1}{2}\,dt$ correctly?

- Oct 19th 2009, 09:52 PMmr fantastic