I saw that substitution should be $u = \arcsin \frac{x}{2^{3/2}}$, but i can't see how can i come to that conclusion. I'd appreciate a little push in the right direction.
Asked
Active
Viewed 168 times
4
-
2$x=\sqrt8\sin(u)$ – Claude Leibovici Feb 09 '17 at 09:04
-
You can do it in two steps: first $x=\sqrt{8}t$, which turns the square root into $\sqrt{8}\sqrt{1-t^2}$; now $u=\arcsin t$ will do. Put the steps together… – egreg Feb 09 '17 at 09:59
1 Answers
6
If we were to substitute $f(u)=x$, well, we want to express $$\sqrt{8-f(u)^2}$$ in some nice kind of function. note that we have to insure that $$8-f(u)^2=g(u)^2 \iff f(u)^2+g(u)^2=8$$ For functions $f,g$ whose integral and derivatives are simple and well known, and $f^{-1}$, the inverse of $f$ exists on the domain of $x$.
The simplest of this is $f(u)=2\sqrt{2} \sin u$, $g(u)=2\sqrt{2} \cos u$ .
So since we have substituted $f(u)=2\sqrt{2} \sin u=x$, this implies that $$u= \arcsin \frac{x}{2^{3/2}}$$ And we see that the integrate evaluates nicely. This is why we use this substitution.
S.C.B.
- 22,768
-
1I didn't pay attention to the domain of the $\sqrt{8 - x^2}$. Now i see how it works. – MathsLearner Feb 09 '17 at 09:33
-
-