1. Integral 2

Calculate:

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

2. Originally Posted by Apprentice123
Calculate:

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

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

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

Solving this we get

$-1
.......

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

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

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

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

Solving this we get

$-1
.......
*Ahem* Beat you to it (which is why the reply was deleted).

7. I use that formula?

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

How to solve?

8. Originally Posted by Apprentice123
I use that formula?

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

How to solve?
$\int \frac{2}{\sqrt{2+x-x^2}} \ \mathrm{d}x = 2 \int \frac{1}{\sqrt{2+x-x^2}} \ \mathrm{d}x =$ $2 \int \frac{1}{\sqrt{\frac{9}{4} - \left(x-\frac12\right)^2}} \ \mathrm{d}x$ $= 2 \int \frac{1}{\sqrt{\left(\frac{3}{2}\right)^2 - \left(x-\frac12\right)^2}} \ \mathrm{d}x$

$2 \int \frac{1}{\sqrt{\left(\frac{3}{2}\right)^2 - \left(x-\frac12\right)^2}} \ \mathrm{d}x$ is in the form $\int \frac{1}{\sqrt{a^2 - u^2}} \ \mathrm{d}u$ so can you continue?

EDIT: Use this formula - $\int \frac{1}{\sqrt{a^2-u^2}} \ \mathrm{d}u = \mathrm{arcsin}\left(\frac{u}{a}\right)$

9. This way?

$\frac{1}{2}arcsin(\frac{(x-\frac{1}{2})}{\frac{3}{2}})$

10. Originally Posted by Apprentice123
This way?

$\frac{1}{2}arcsin(\frac{(x-\frac{1}{2})}{\frac{3}{2}})$
You forgot the $2$ which was outside the integral (the numerator).

Also, the general rule is $\int \frac{1}{\sqrt{a^2-u^2}} \ \mathrm{d}u = \mathrm{arcsin}\left(\frac{u}{a}\right)$ hence you do not need $\frac12$ at the start.

$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)}
$2\left[\mathrm{arcsin}\left(\frac{2\left (x-\frac12\right)}{3}