# Got this one in my test today...

• Apr 16th 2008, 08:09 AM
janvdl
Got 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)

$\displaystyle \int \frac{1 - t}{\sqrt{9 - t^2}}$
• Apr 16th 2008, 08:26 AM
polymerase
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 AM
Isomorphism
Hello janvdl,

Substitute $\displaystyle t = 3\sin x$

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

• Apr 16th 2008, 08:30 AM
janvdl
Quote:

Originally Posted by polymerase
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 $\displaystyle \int \frac{1-t}{\sqrt{9 - t^2}}=\sqrt{9-t^2}-\arcsin\left(\frac{\sqrt{9-t^2}}{3}\right)+C$

Hehe okay, i got it wrong then... Thanks! ;)
• Apr 16th 2008, 08:32 AM
janvdl
Quote:

Originally Posted by Isomorphism
Hello janvdl,

Substitute $\displaystyle t = 3\sin x$

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

Thanks for your reply, but I don't understand why you substituted using Sin(x). Maybe that is a more advanced method?
• Apr 16th 2008, 08:36 AM
polymerase
Quote:

Originally Posted by janvdl
Thanks for your reply, but I don't understand why you substituted using Sin(x). Maybe that is a more advanced method?

That's trig. substitution. By letting $\displaystyle t=3\sin x$, you end up with $\displaystyle \sqrt{9-9\sin^2 x}=3\sqrt{1-\sin^2 x}=3\sqrt{\cos^2 x}=3\cos x$ Get it?
• Apr 16th 2008, 08:37 AM
janvdl
Quote:

Originally Posted by polymerase
That's trig. substitution. By letting $\displaystyle t=3\sin x$, you end up with $\displaystyle \sqrt{9-9\sin^2 x}=3\sqrt{1-\sin^2 x}=3\sqrt{\cos^2 x}=3\cos x$ Get it?

Got it :D Thanks.
• Apr 16th 2008, 08:41 AM
Isomorphism
Quote:

Originally Posted by janvdl
Thanks for your reply, but I don't understand why you substituted using Sin(x). Maybe that is a more advanced method?

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 AM
janvdl
Right, I think I got it now ;) Thanks again guys! (Handshake)