Hello everyone (Hi)

I got this one in my test today and would like to know what the answer should have been, please. (EDIT: I used partial integration on it)

Thanks in advance.

$\displaystyle \int \frac{1 - t}{\sqrt{9 - t^2}}$

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- Apr 16th 2008, 08:09 AMjanvdlGot this one in my test today...
Hello everyone (Hi)

I got this one in my test today and would like to know what the answer should have been, please. (EDIT: I used partial integration on it)

Thanks in advance.

$\displaystyle \int \frac{1 - t}{\sqrt{9 - t^2}}$ - Apr 16th 2008, 08:26 AMpolymerase
make the substitution $\displaystyle u^2=9-t^2$

You then end up with this, $\displaystyle -\int \frac{1-\sqrt{9-u^2}}{\sqrt{9-u^2}}=u-\arcsin \left(\frac{u}{3}\right)+C $

Thus $\displaystyle \int \frac{1-t}{\sqrt{9 - t^2}}=\sqrt{9-t^2}-\arcsin\left(\frac{\sqrt{9-t^2}}{3}\right)+C$ - Apr 16th 2008, 08:28 AMIsomorphism
Hello janvdl,

Substitute $\displaystyle t = 3\sin x$

We get $\displaystyle \displaystyle \int (1 - 3\sin x) \, dx = x + 3\cos x + C = \sin^{-1} \frac{t}{3} + 3\sqrt{1 - t^2}+C$

Whats your method? - Apr 16th 2008, 08:30 AMjanvdl
- Apr 16th 2008, 08:32 AMjanvdl
- Apr 16th 2008, 08:36 AMpolymerase
- Apr 16th 2008, 08:37 AMjanvdl
- Apr 16th 2008, 08:41 AMIsomorphism
Not really,if you did learn substitution, I think you should be familiar with this.

Whenever you see $\displaystyle \sqrt{a^2 - x^2}$ there is an itching to get rid of the square root. And the standard substitution is $\displaystyle x = a\sin u$. $\displaystyle \sqrt{a^2 - x^2}$ neatly simplifies to $\displaystyle a\cos u$. Moreover $\displaystyle dx = a\cos u \, du$. Everything neatly becomes trigonometric (Wink) - Apr 16th 2008, 08:42 AMjanvdl
Right, I think I got it now ;) Thanks again guys! (Handshake)