How can I evaluate this integral?
$$ \int_2^3 \frac{1}{\sqrt{x^2-x-2}} dx$$
I know that this integral is convergent but I don't know how to evaluate it.
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The first thing you think of should be completing the square. That gives you $(x + \cdots\cdots)^2 + \cdots\cdots = u^2 + \cdots\cdots$. – Michael Hardy Nov 22 '11 at 20:54
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$$I = \displaystyle \int_2^3 \frac{dx}{\sqrt{x^2 - x -2}} = \displaystyle \int_2^3 \frac{dx}{\sqrt{ \left(x - \frac12 \right)^2 - \left(\frac{3}{2} \right)^2}}$$ Now let $\left(x - \frac12 \right) = \frac32 \sec(\theta)$. This gives us $dx = \frac32 \sec(\theta) \tan(\theta) d \theta$. Hence, we get $$I = \int_{0}^{\theta_1} \sec(\theta) d \theta = \log \left( \sec(\theta_1) + \tan(\theta_1) \right)$$ where $\cos(\theta_1) = \frac35$. Hence, we get $$I = \log \left(\frac53 + \frac43 \right) = \log(3)$$
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Put $(x^2-x)= (x-1/2)^2-1/4$ to obtain $\int_2^3 \dfrac{1}{\sqrt{x^2-x-2}}dx=\int_2^3 \dfrac{1}{\sqrt {(x-1/2)^2-9/4}}dx$ and do the change of variable $u=x-1/2$ then the integral becomes $\int_{3/2}^{5/2} \dfrac{1}{\sqrt {u^2-9/4}} du=...$ the rest is an exercise.
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Since this is very likely a homework problem,
HINT: $x^2-x-2 = (x+1)(x-2)$. Thus $\int_2^3 \frac{\mathrm{d} x}{\sqrt{(x+1)(x-2)}} \stackrel{y=x+2}{=} \int_0^1 \frac{\mathrm{d} y}{\sqrt{(y+3) y}} \stackrel{y=u^2}{=} \int_0^1 \frac{2 \mathrm{d} u}{\sqrt{3+u^2}}$.
Sasha
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1Do you mean how to find the closed form value of the integral. This is done with substitution $u=\sqrt{3} \sinh t$. Then $\mathrm{d} u = \sqrt{3} \cosh(t) \mathrm{d} t$ and $\sqrt{3 + u^2} = \sqrt{3} \cosh(t)$. Can you finish it ? – Sasha Nov 22 '11 at 19:56