1. ## Integral 2

Calculate:

$\displaystyle \int \frac{2dx}{\sqrt{2+x-x^2}}$

2. Originally Posted by Apprentice123
Calculate:

$\displaystyle \int \frac{2dx}{\sqrt{2+x-x^2}}$
$\displaystyle 2+x-x^2=2-(x^2-x)=2-\left(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right)=\frac{9}{4}-\left(x-\frac{1}{2}\right)^2$

Let $\displaystyle x-\frac{1}{2}=\frac{3}{2}\sin(\theta)$

3. Originally Posted by mr fantastic
$\displaystyle 2 + x - x^2 < 0$ for all real values of x so the integrand makes no sense.
$\displaystyle 2+x-x^2>0$

Solving this we get

$\displaystyle -1<x<2$
.......

4. Originally Posted by Mathstud28
$\displaystyle 2+x-x^2=2-(x^2-x)=2-\left(\left(x-\frac{1}{2}\right)^2-\frac{1}{4}\right)=\frac{9}{4}-\left(x-\frac{1}{2}\right)^2$

Let $\displaystyle x-\frac{1}{2}=\sin(\theta)$
you mean $\displaystyle x - \frac 12 = \frac 32 \sin \theta$

5. Originally Posted by Jhevon
you mean $\displaystyle x - \frac 12 = \frac 32 \sin \theta$
Yeah, thank you...forgot to type that part

6. Originally Posted by Mathstud28
$\displaystyle 2+x-x^2>0$

Solving this we get

$\displaystyle -1<x<2$
.......
*Ahem* Beat you to it (which is why the reply was deleted).

7. I use that formula?

$\displaystyle \int \frac{du}{\sqrt{u^2+a^2}}=ln(u+\sqrt{u^2+a^2})$

How to solve?

8. Originally Posted by Apprentice123
I use that formula?

$\displaystyle \int \frac{du}{\sqrt{u^2+a^2}}=ln(u+\sqrt{u^2+a^2})$

How to solve?
$\displaystyle \int \frac{2}{\sqrt{2+x-x^2}} \ \mathrm{d}x = 2 \int \frac{1}{\sqrt{2+x-x^2}} \ \mathrm{d}x =$ $\displaystyle 2 \int \frac{1}{\sqrt{\frac{9}{4} - \left(x-\frac12\right)^2}} \ \mathrm{d}x $$\displaystyle = 2 \int \frac{1}{\sqrt{\left(\frac{3}{2}\right)^2 - \left(x-\frac12\right)^2}} \ \mathrm{d}x \displaystyle 2 \int \frac{1}{\sqrt{\left(\frac{3}{2}\right)^2 - \left(x-\frac12\right)^2}} \ \mathrm{d}x is in the form \displaystyle \int \frac{1}{\sqrt{a^2 - u^2}} \ \mathrm{d}u so can you continue? EDIT: Use this formula - \displaystyle \int \frac{1}{\sqrt{a^2-u^2}} \ \mathrm{d}u = \mathrm{arcsin}\left(\frac{u}{a}\right) 9. This way? \displaystyle \frac{1}{2}arcsin(\frac{(x-\frac{1}{2})}{\frac{3}{2}}) 10. Originally Posted by Apprentice123 This way? \displaystyle \frac{1}{2}arcsin(\frac{(x-\frac{1}{2})}{\frac{3}{2}}) You forgot the \displaystyle 2 which was outside the integral (the numerator). Also, the general rule is \displaystyle \int \frac{1}{\sqrt{a^2-u^2}} \ \mathrm{d}u = \mathrm{arcsin}\left(\frac{u}{a}\right) hence you do not need \displaystyle \frac12 at the start. Therefore the answer is: \displaystyle 2 \int \frac{1}{\sqrt{\left(\frac{3}{2}\right)^2 - \left(x-\frac12\right)^2}} \ \mathrm{d}x = 2\left[\mathrm{arcsin}\left(\frac{\left(x-\frac12\right)}{\left(\frac32\right)} \right) \right]=$$\displaystyle 2\left[\mathrm{arcsin}\left(\frac{2\left (x-\frac12\right)}{3} \right)\right]= 2\left[\mathrm{arcsin}\left(\frac{\left (2x-1\right)}{3} \right)\right] = 2\mathrm{arcsin}\left(\frac{2x-1}{3} \right)$

11. thank you