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I can not get the correct answer.

$$\int \frac {dx}{\sqrt {x^2 + 16}}$$

$x = 4 \tan \theta$, $dx = 4\sec^2 \theta$

$$\int \frac {dx}{\sqrt {16 \sec^2 \theta}}$$

$$\int \frac {4 \sec^ 2 \theta}{\sqrt {16 \sec^2 \theta}}$$

$$\int \frac {4 \sec^ 2 \theta}{4 \sec \theta}$$

$$\int \sec \theta$$

$$\ln| \sec \theta + \tan \theta|$$

Then I solve for $\theta$:

$x = 4 \tan \theta$

$x/4 = \tan \theta$

$\arctan (\frac{x}{4}) = \theta$

$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tan (\arctan (\tfrac{x}{4}))|$$

$$\ln| \sec (\arctan (\tfrac{x}{4})) + \tfrac{x}{4}))| + c$$

This is wrong and I do not know why.

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    $\begingroup$ It looks good. All that's left to do is simplify $\sec(\arctan(x/4))$. $\endgroup$ – David Mitra Jun 4 '12 at 15:43
  • $\begingroup$ Looks pretty good to me. But $\sec^2\theta=1+\tan^2\theta=1+(x/4)^2$, so $\sec\theta=\sqrt{1+(x/4)^2}=\frac{1}{4}\sqrt{16+x^2}$. (I am being sloppy and not worrying about signs. Should!) Can simplify further, using logarithm laws, and changing the arbitrary constant. $\endgroup$ – André Nicolas Jun 4 '12 at 15:43
  • $\begingroup$ Yes, you just have it in a slightly different form than would normally be used. Also, there are some slight mistakes in your notation throughout the process. $\endgroup$ – process91 Jun 4 '12 at 15:45
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What you have done is correct! Note that whenever you have inverse trigonometric expressions you can express your answer in more than one way! Your answer can be expressed in a different way (without the trigonometric and inverse trigonometric functions) as shown below.

We will prove that $$\sec \left( \arctan \left( \dfrac{x}4 \right) \right) = \sqrt{1 + \left(\dfrac{x}{4} \right)^2}$$

Hence, your answer $$\ln \left \lvert \dfrac{x}4 + \sec \left(\arctan \left( \dfrac{x} 4\right) \right) \right \rvert + c$$ can be rewritten as $$\ln \left \lvert \dfrac{x}4 + \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \right \rvert + c$$ Note that $$\theta = \arctan\left( \dfrac{x}4 \right) \implies \tan( \theta) = \dfrac{x}4 \implies \tan^2(\theta) = \dfrac{x^2}{16} \implies 1 + \tan^2(\theta) = 1+\dfrac{x^2}{16}$$ Hence, we get that $$\sec^2(\theta) = 1+ \left(\dfrac{x}{4} \right)^2 \implies \sec (\theta) = \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \implies \sec \left(\arctan\left( \dfrac{x}4 \right) \right) = \sqrt{1+ \left(\dfrac{x}{4} \right)^2}$$ Hence, you can rewrite your answer as $$\ln \left \lvert \dfrac{x}4 + \sqrt{1+ \left(\dfrac{x}{4} \right)^2} \right \rvert + c$$

Also, you have been a bit sloppy with some notations in your argument.

For instance, when you substitute $x = 4 \tan (\theta)$, $$\dfrac{dx}{\sqrt{x^2+16}} \text{ should immediately become }\dfrac{4 \sec^2(\theta)}{\sqrt{16 \sec^2(\theta)}} d \theta$$

Also, you need to carry the $d \theta$ throughout the answer under the integral.

Writing just $\displaystyle \int\sec(\theta)$ or $\displaystyle \int\dfrac{4 \sec^2(\theta)}{\sqrt{16 \sec^2(\theta)}}$ without the $d \theta$ is notationally incorrect.

Anyway, I am happy that you are slowly getting a hang of these!

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Cosmetically nicer taking $$ x = 4 \sinh t \; , $$ so that $$ dx = 4 \cosh t \; dt $$ The quadratic formula is enought ot give us $$ t = \log \left( \frac{x + \sqrt{x^2 + 16}}{4} \right) = \log \left( x + \sqrt{x^2 + 16} \right) - \log 4 $$

Then we get $$ \int \frac{dx}{x + \sqrt{x^2 + 16} } = \int 1 dt = t + C = \log \left( x + \sqrt{x^2 + 16} \right) - \log 4 + C \; , $$ or $$ \int \frac{dx}{x + \sqrt{x^2 + 16} } = \log \left( x + \sqrt{x^2 + 16} \right) + C_2 \; . $$

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$$\int \frac {dx}{\sqrt {x^2 + 16}}$$

Let $$t=\sqrt {x^2+16}$$ this will change the integral into $$ \int \frac {dt}{\sqrt {t^2 - 16}}$$

$$t=4\sec (\theta)$$ changes the integral into $$\int \sec \theta d \theta =ln| \sec \theta + \tan \theta|+c$$

$$=\ln (t/4 + \frac {\sqrt {t^2-16}}{4} )+c$$ $$= \ln ( \sqrt \frac {x^2+16}{4}+x/4 )+c$$

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