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Evaluate the definite integral: $$\int_{1/3}^{\sqrt{2}/3} \dfrac{1}{x\sqrt{9x^2-1}}{dx}$$ So,

I am to use $u$-substitution, and immediately, it would appear that perhaps the integral may be some function of x in the form $\sqrt{x}, \ln{x},$ or possibly some inverse trigonometric function.

So I feel like the most progress I've made was in taking $u = 9x^2 -1 \implies du = 18x {\ dx}$. The concept of this method is very strange to me, as well (though seemingly I know how to apply it, in most cases), because, conceptually, taking and moving the $dx$ in place of the $du$ arithmetically seems a little bit like having $1+1 = 11$. Perhaps I just don't understand exactly what it's all implying?

But, aside from not really understanding why $u$-substitution works, I really don't know what to take for $u$ in order to evaluate this integral using $u$-substitution.

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3 Answers 3

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Try substituting $u=\sqrt{9x^2-1}$. Then $du=\frac{9x}{\sqrt{9x^2-1}}dx=\frac{9x^2}{x\sqrt{9x^2-1}}dx$

Notice that this implies $$\frac{dx}{x\sqrt{9x^2-1}}=\frac{du}{9x^2}=\frac{du}{1+u^2}$$

Thus,

$$\int_{1/3}^{\sqrt{2}/3} \frac{dx}{x\sqrt{9x^2-1}}=\int_{0}^{1} \frac{du}{1+u^2}=\arctan(1)-\arctan(0)=\pi/4$$


NOTE: Primer on Integration by Substitution

The fundamental theorem of calculus states that if

$$F(x)=\int_a^x f(t) dt$$

then

$$F'(x) = f(x)$$

Now, let $G$ be defined as the integral (This is a substitution $t=g(u)$ and we will show that $F=G$ to complete the primer.)

$$G(x)= \int_{g^{-1}(a)}^{g^{-1}(x)}f(g(u))\frac{dg(u)}{du}du$$

From the chain rule we have

$$\begin{align} G'(x) &=G'(g^{-1}(x))\frac{dg^{-1}(x)}{dx}\\\\ &=\left(f(g(g^{-1}(x)))\frac{dg(g^{-1}(x))}{dg^{-1}(x)}\right)\frac{dg^{-1}(x)}{dx}\\\\ &=f(x) \frac{dx}{dg^{-1}(x)}\frac{dg^{-1}(x)}{dx}\\\\ &=f(x) \end{align}$$

Thus, we have that $F'(x) = G'(x)$. Since $F(a)=G(a)$, then $F(x) = G(x)$.

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  • $\begingroup$ Could you maybe explain the intuition behind $u$-substitution for me? I find the whole method strange (though I do recognise that it works as intended). It seems a bit of a mystical thing (and I'm really not a fan of magic, if you know what I mean). $\endgroup$
    – alxmke
    Commented May 7, 2015 at 1:28
  • $\begingroup$ Did you see my post? $\endgroup$
    – alxmke
    Commented May 7, 2015 at 3:18
  • $\begingroup$ @Alex Yes and I added a primer to the answer. I really hope that it helps. I just want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Commented May 7, 2015 at 3:20
  • $\begingroup$ Apologies, I hadn't noticed. I'm reading it now. Thanks. $\endgroup$
    – alxmke
    Commented May 7, 2015 at 3:20
  • $\begingroup$ No apology necessary. And you're welcome. My pleasure. $\endgroup$
    – Mark Viola
    Commented May 7, 2015 at 3:23
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It's an Arcsecant. We can rewrite that integral as $$\int_{\frac{1}{3}}^{\frac{\sqrt{2}}{3}} \frac{1}{x\sqrt{(\frac{x}{\frac{1}{3}})^2-1}}$$ Which is $$\frac{1}{3}\sec^{-1}x |_{\frac{1}{3}}^{\frac{\sqrt{2}}{3}}$$ Hope that helps.

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  • $\begingroup$ I thought that the OP requested a "u" substitution. $\endgroup$
    – Mark Viola
    Commented May 7, 2015 at 0:28
  • $\begingroup$ Let $u=\text{arcsec}(3x)$. $\endgroup$ Commented May 7, 2015 at 2:54
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Hint: On one hand, $\big(9x^2-1\big)'=18x$. On the other hand, $\dfrac1{x\sqrt{9x^2-1}}=\dfrac x{x^2\sqrt{9x^2-1}}=$

$=\dfrac{\bigg[\dfrac{\big(18x\big)}{18}\bigg]}{\bigg[\dfrac{~\big(9x^2-1\big)+1}9\bigg]\cdot\sqrt{9x^2-1}~}~.~$ Can you take it from here ? ;-$)$

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