Solving an integral using formula #30 from the integral table

[SilverCrow]

Hi,

The problem is the integral of sqrt(16x-2x^2-23):

\(\displaystyle \mbox{Find }\, \int\, \sqrt{\displaystyle 16x\, -\, 2x^2\, -\, 23\,}\, dx\)

I know that I have to complete the square first and then use the formula:

\(\displaystyle \mbox{#30: }\, \int\, \sqrt{\displaystyle a^2\, -\, u^2\,}\, dx\, \) \(\displaystyle =\, \left(\dfrac{u}{2}\right)\,\) \(\displaystyle \sqrt{\displaystyle a^2\, -\, u^2\, }\, \) \(\displaystyle +\, \left(\dfrac{a^2}{u^2}\right)\, \sin^{-1}\left(\dfrac{a^2}{2}\right)\, +\, C\)

I completed the square and have -2(x-4)^2+9.

 

[stapel]

\(\displaystyle \mbox{Find }\, \int\, \sqrt{\displaystyle 16x\, -\, 2x^2\, -\, 23\,}\, dx\)

I know that I have to complete the square first and then use the formula:

\(\displaystyle \mbox{#30: }\, \int\, \sqrt{\displaystyle a^2\, -\, u^2\,}\, dx\, \) \(\displaystyle =\, \left(\dfrac{u}{2}\right)\,\) \(\displaystyle \sqrt{\displaystyle a^2\, -\, u^2\, }\, \) \(\displaystyle +\, \left(\dfrac{a^2}{u^2}\right)\, \sin^{-1}\left(\dfrac{a^2}{2}\right)\, +\, C\)

I completed the square and have -2(x-4)^2+9.

Okay. Now try to fit what you've got into the formula they told you to use. For instance, one could restate the completed-square form as follows:

. . . . .\(\displaystyle 9\, -\, 2(x\, -\, 4)^2\)

Then:

. . . . .\(\displaystyle \mbox{Let }\, u\, =\, \sqrt{2\,}(x\, -\, 4),\, \mbox{ so }\, du\, =\, \sqrt{2\,}\, dx,\,\mbox{ and }\, \dfrac{du}{\sqrt{2\,}}\, =\, dx\)

. . . . .\(\displaystyle \mbox{Then }\, \left(9\, -\, 2(x\, -\, 4)^2\right)\, dx\, =\, \left(3^2\, -\, u^2\right)\, \dfrac{du}{\sqrt{2\,}}\)

Then what?

If you get stuck, please reply showing all of your steps. Thank you!