Trigonometric substitution

**emurphy** · October 19th 2009, 07:38 PM

Evaluate the indefinite integral.

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

**redsoxfan325** · October 19th 2009, 07:46 PM

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 $t=\sin^{-1}\left(\frac{x-8}{8}\right)$ .

**emurphy** · October 19th 2009, 07:57 PM

I got this to be my answer

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

**mr fantastic** · October 19th 2009, 09:15 PM

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

**emurphy** · October 19th 2009, 09:51 PM

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?

**mr fantastic** · October 19th 2009, 09:52 PM

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.

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