An integral with the square root of a function

[suzanka]

I'm trying to solve the problem: Integral of \(\displaystyle 1 / sqrt(3+2x-x^2)\) but I can't get to the answer given in the textbook. The polynom equals to \(\displaystyle -(x-3)(x+1)\) so I applied the Euler substitution \(\displaystyle sqrt (3+2x-x^2) = (x+1)t\). I got \(\displaystyle x = (3-t^2) / (t^2+1)\) and \(\displaystyle dx = -8t / ( (t^2+1)^2 )\). The answer I get is \(\displaystyle -2arctan(t) = -2arctan( sqrt(3+2x-x^2) / (x-1) ) + C\) and the textbook says it's \(\displaystyle -arcsin( (1-x) / 2 ) + C\). I tried solving it a few times but I can't find where my error is. Help appreciated

 

[Deleted member 4993]

I'm trying to solve the problem: Integral of \(\displaystyle 1 / sqrt(3+2x-x^2)\) but I can't get to the answer given in the textbook. The polynom equals to \(\displaystyle -(x-3)(x+1)\) so I applied the Euler substitution \(\displaystyle sqrt (3+2x-x^2) = (x+1)t\). I got \(\displaystyle x = (3-t^2) / (t^2+1)\) and \(\displaystyle dx = -8t / ( (t^2+1)^2 )\). The answer I get is \(\displaystyle -2arctan(t) = -2arctan( sqrt(3+2x-x^2) / (x-1) ) + C\) and the textbook says it's \(\displaystyle -arcsin( (1-x) / 2 ) + C\). I tried solving it a few times but I can't find where my error is. Help appreciated

\(\displaystyle \displaystyle \int \frac{dx}{\sqrt{3+2x-x^2}} \)

\(\displaystyle \displaystyle = \int \frac{dx}{\sqrt{4 - (x-1)^2}} \)

substitute

x -1 = 2 sin(Θ)

dx = 2 cos(Θ) dΘ

The integral becomes:

\(\displaystyle \displaystyle = \int \frac{2*cos(\theta)d\theta}{2*cos(\theta)}\)

\(\displaystyle \displaystyle = \theta \ + \ C\)

Now back substitute...

 

[suzanka]

Thank you very much, I had forgotten about trigonometric substitution.