\[ 2 \arcsin (\sqrt {x})-\arcsin (2x-1)= \frac {\pi }{2} \]
Consider the indefinite integral:
\[ I = \int \frac {1}{\sqrt {x-x^2}} \d x \]
One way to compute \(I\) is to set \(u=2x-1\) and write the integral \(I\) as an integral with respect
to \(u\):
\[ \int \frac {1}{\sqrt {x-x^2}} \d x = \int \frac {1}{ \answer {\sqrt {1-u^2}}} \d u \]
You will need to do some algebra!
Evaluate this integral in \(u\) and reverse the substitution to write \(\int \frac {1}{\sqrt {x-x^2}} \d x\) in terms of
\(x\):
(Use \(+C\) for the constant of integration)
\[ I = \answer {\arcsin (2x-1)+C} \]
You may find the result \(\int \frac {1}{\sqrt {a^2-x^2}} \d x = \arcsin \left (\frac {x}{a}\right )\) helpful.
Another way to compute \(I\) is to set \(v=\sqrt {x}\) and write \(I\) as an integral with respect to
\(v\).
\[ \int \frac {1}{\sqrt {x-x^2}} \d x = \int \frac {2}{ \answer {\sqrt {1-v^2}}} \d v \]
You will need to do some algebra! Note that by definition that \(v\geq 0\), so \(\sqrt {v^2} = v\) (recall that
usually \(\sqrt {v^2} = |v|\) because by convention, the square root is positive!)
If you work the substitution correctly, you should obtain:
\[ \int \frac {1}{\sqrt {x-x^2}} \d x = \int \frac {\answer {2v}}{ \answer {\sqrt {v^2-v^4}}} \d v \]
Factor a \(v^2\) out of the denominator and simplify!
Evaluate this integral in \(v\) and reverse the substitution to write \(\int \frac {1}{\sqrt {x-x^2}} \d x\) in terms of
\(x\):
(Use \(+C\) for the constant of integration)
\[ I = \answer {2\arcsin (\sqrt {x})+C} \]
You have now computed \( \int \frac {1}{\sqrt {x-x^2}} \d x\) two different ways! The antiderivatives \(2\arcsin (\sqrt {x})\) and \(\arcsin (2x-1)\) above can differ only by a constant.
Using this observation:
\[ 2\arcsin (\sqrt {x})-\arcsin (2x-1) = \answer {\frac {\pi }{2}} \]
Since these antiderivatives differ by a constant, we have:
\[ 2\arcsin (\sqrt {x})-\arcsin (2x-1) = C \]
and this constant \(C\) must
be the same no matter what x-value we use (this is precisely what it means for \(C\) to be
a constant!)
Set \(x=1\) above and find \(C\).