When I enter the following Integral problem to Mathematica, $$ \int \frac{1}{\sqrt{a^2-x^2}}\textrm{d}x $$ The answer yielded by Mathematica is $$ \arctan\left({\frac{x}{\sqrt{a^2-x^2}}}\right) $$ Rather than the more reduced result: $$ \arcsin\left(\frac{x}{a}\right). $$ The two results are mathematically equivalent but I think the $\arcsin$ result is more typically considered the answer to the integral. At least, every table of integrals that I own plus a few on-line queries always give the $\arcsin$ answer.
There must be a way in Mathematica to force the $\arcsin$ result as preference to the $\arctan$ equivalent but I have not found it. I tried to use Simplify
and even used Assumptions
on the integral to declare that $a$ is real (thinking that might have been an issue).
Question: What do I do in Mathematica to force the result to be the $\arcsin$ version of the answer?
Statements I have used include:
Integrate[1/Sqrt[a^2-x^2],x]
and,
Integrate[1/Sqrt[a^2-x^2],x]//Simplify
and,
Integrate[1/Sqrt[a^2-x^2],x,Assumptions->Element[a,Reals]]
Updated
I attempted various other statements and in particular, dealing with the variable $a$ being given explicit values. For example,
Integrate[1/Sqrt[a^2-x^2],x,Assumptions->a>0]
Still results in the ArcTan
form of the result. However, if I explicitly specify a positive value such as $a=2$ then the result is the ArcSin
format as shown:
a=2; Integrate[1/Sqrt[a^2-x^2],x]
With result:
$$
\arcsin\left(\frac{x}{2}\right).
$$
Either form has a forbidden zone for $x>a$. For the ArcTan
, you have Complex values and for the ArcSin
you have a situation where the magnitude of the Sin
function is greater than 1.
a = 1; Plot[{Re[ArcSin[x/a]], Im[ArcSin[x/a]], Re[ArcTan[x/Sqrt[a^2-x^2]]], Im[ArcTan[x/Sqrt[a^2-x^2]]]}, {x,-2,2}]
MMA by default considers complex and not just real. $\endgroup$Simplify[1/Sqrt[a^2-x^2]-D[ArcSin[x/a],x],Assumptions->a>0]
gives zero. But I could not make Intergate gives the simplified result with assumptions that a>0. Maple can do it.int( 1/(sqrt(a^2-x^2)),x) assuming a>0;
givesarcsin(x/a)
may be someone else can figure how to make Integrate do it. !Mathematica graphics $\endgroup$x==a
.ArcSin[x/a] /. x -> 2 a // N
yields1.5708 - 1.31696 I
, perfectly sensible. $\endgroup$