# More help in integrating

1. Sep 17, 2007

### rock.freak667

[SOLVED] More help in integrating

1. The problem statement, all variables and given/known data

Find $$a,b,n$$ such that $$a,b$$ are rational and n is not an element of Z. a>0,b<0

$$\int_{0}^{1} \frac{3x}{\sqrt{1+6x-3x^2}} = \frac{\pi}{a^n} + b$$

2. Relevant equations
$$\int \frac{1}{\sqrt{A^2-X^2}} dx= \arcsin{\frac{X}{A}+K$$

$$\int \frac{x}{R} = \frac{R}{a} -\frac{b}{2a}\int\frac{dx}{R} where R=\sqrt{ax^2+bx+c$$(Note: this really can't be used as working must be shown)

3. The attempt at a solution

$$\int_{0}^{1} \frac{3x}{\sqrt{1+6x-3x^2}} dx = \int_0^{1} \frac{3x}{\sqrt{4-3(x-1)^2} }dx$$

$$= \int_{0}^{1} \frac{3x}{\sqrt{(2)^2-(\sqrt{3(x-1)}})^2} dx$$

Now I can't really see how to get rid of the "x" in the numerator...and way i can get rid of it?(by a nice substitution?

Last edited: Sep 17, 2007
2. Sep 17, 2007

### rock.freak667

Well a bit more progress has been made:

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

Making the entire integral become:

$$\int\frac{2\sqrt{3}+3}{\sqrt{4-4sin^2\theta}} \frac{2cos\theta}{\sqrt{3}} d\theta$$
simplifying to give

$$2\int sin\theta d\theta +\int\sqrt{3} d\theta$$

$$= -2cos\theta + \sqrt{3}\theta$$

$$= \left[-2cos({sin^{-1}(\frac{\sqrt{3}(x-1)}{2})) + \sqrt{3}sin^{-1}(\frac{\sqrt{3}(x-1)}{2}\right]_{0}^{1}$$

giving finally
$$-2+\frac{\pi}{2\sqrt{3}}$$ which is not the form I need it in