MHB Evaluate $\int_{1}^{2}\dfrac {dx}{(x^2-2x+4)^{3/2}}$

[Albert1]

$ \begin{align*} \int_{1}^{2}\dfrac {dx}{(x^2-2x+4)^{3/2}}\end{align*}$

 

[M R]

Re: evaluate integral-02

As a first step I would complete the square. :)

 

[MarkFL]

Re: evaluate integral-02

As a first step I would complete the square. :)

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[Prove It]

Re: evaluate integral-02

$\int_{1}^{2}\dfrac {dx}{(x^2-2x+4)^{3/2}}$

[math]\displaystyle \begin{align*} \int_1^2{\frac{dx}{\left( x^2 - 2x + 4 \right) ^{\frac{3}{2}} }} &= \int_1^2{\frac{dx}{\left[ \left( x - 1 \right) ^2 + 3 \right] ^{\frac{3}{2}} }} \end{align*}[/math]

Now make the substitution [math]\displaystyle \begin{align*} x - 1 = \sqrt{3}\,\tan{(\theta)} \implies dx = \sqrt{3}\,\sec^2{(\theta)}\,d\theta \end{align*}[/math] and noting that when [math]\displaystyle \begin{align*} x = 1, \, \theta = 0 \end{align*}[/math] and when [math]\displaystyle \begin{align*} x = 2, \, \theta = \frac{\pi}{6} \end{align*}[/math], then the integral becomes

[math]\displaystyle \begin{align*} \int_1^2{\frac{dx}{\left[ \left( x - 1 \right) ^2 + 3 \right] ^{\frac{3}{2}}}} &= \int_0^{\frac{\pi}{6}}{\frac{\sqrt{3}\,\sec^2{( \theta )} \, d\theta }{\left[ \left( \sqrt{3}\, \tan{(\theta)} \right) ^2 + 3 \right] ^{\frac{3}{2}}}} \\ &= \int_0^{\frac{\pi}{6}}{\frac{\sqrt{3}\,\sec^2{( \theta )}\,d\theta}{\left[ 3\sec^2{(\theta)} \right] ^{\frac{3}{2}} }} \\ &= \int_0^{\frac{\pi}{6}}{ \frac{\sqrt{3}\,\sec^2{( \theta )} \, d\theta }{3\sqrt{3}\,\sec^3{( \theta )}} } \\ &= \int_0^{\frac{\pi}{6}}{\frac{d\theta }{ 3\sec{( \theta )} }} \\ &= \frac{1}{3} \int_0^{\frac{\pi}{6}}{\cos{( \theta )} \, d\theta } \\ &= \frac{1}{3} \left[ \sin{( \theta )} \right] _0 ^{\frac{\pi}{6}} \\ &= \frac{1}{3} \left( \frac{1}{2} - 0 \right) \\ &= \frac{1}{6} \end{align*}[/math]

 

[Albert1]

Re: evaluate integral-02

[math]\displaystyle \begin{align*} \int_1^2{\frac{dx}{\left( x^2 - 2x + 4 \right) ^{\frac{3}{2}} }} &= \int_1^2{\frac{dx}{\left[ \left( x - 1 \right) ^2 + 3 \right] ^{\frac{3}{2}} }} \end{align*}[/math]
Now make the substitution [math]\displaystyle \begin{align*} x - 1 = \sqrt{3}\,\tan{(\theta)} \implies dx = \sqrt{3}\,\sec^2{(\theta)}\,d\theta \end{align*}[/math] and noting that when [math]\displaystyle \begin{align*} x = 1, \, \theta = 0 \end{align*}[/math] and when [math]\displaystyle \begin{align*} x = 2, \, \theta = \frac{\pi}{6} \end{align*}[/math]
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perfect ! you got it,your solution is really (Cool)