definite integration!

**lawochekel** · April 16th 2012, 02:17 AM

$\int\frac{1}{\sqrt{5x^2-2x-4}}dx$ x=1 and x=0

pls i need help on this problem above, thanks

**Prove It** · April 16th 2012, 02:37 AM

Originally Posted by lawochekel

$\int\frac{1}{\sqrt{5x^2-2x-4}}dx$ x=1 and x=0

pls i need help on this problem above, thanks

$\displaystyle \begin{align*} \int_0^1{\frac{1}{\sqrt{5x^2 - 2x - 4}}\,dx} &= \int_0^1{\frac{1}{\sqrt{5\left(x^2 - \frac{2}{5}x - \frac{4}{5}\right)}}\,dx} \\ &= \int_0^1{\frac{1}{\sqrt{5 \left[ x^2 - \frac{2}{5}x + \left(-\frac{1}{5}\right)^2 - \left(-\frac{1}{5}\right)^2 - \frac{4}{5} \right]} }\,dx} \\ &= \int_0^1{\frac{1}{\sqrt{5\left[\left(x - \frac{1}{5}\right)^2 - \frac{1}{25} - \frac{20}{25} \right]}}\,dx} \\ &= \int_0^1{\frac{1}{\sqrt{5\left[\left(x - \frac{1}{5}\right)^2 - \frac{21}{25}\right]}}\,dx} \\ &= \frac{1}{\sqrt{5}}\int_0^1{\frac{1}{\sqrt{\left(x-\frac{1}{5}\right)^2-\frac{21}{25}}}\,dx} \end{align*}$

Now make the substitution $\displaystyle \begin{align*} x - \frac{1}{5} = \frac{\sqrt{21}}{5}\cosh{t} \implies dx = \frac{\sqrt{21}}{5}\sinh{t}\,dt \end{align*}$ .

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