Hello everyone
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.
$\displaystyle \int \frac{1 - t}{\sqrt{9 - t^2}}$
Hello everyone
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.
$\displaystyle \int \frac{1 - t}{\sqrt{9 - t^2}}$
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 $\displaystyle \int \frac{1-t}{\sqrt{9 - t^2}}=\sqrt{9-t^2}-\arcsin\left(\frac{\sqrt{9-t^2}}{3}\right)+C$
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