4
$\begingroup$

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.

  • 0
    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.2012-06-04

3 Answers 3

8

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!

1

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 \; . $

0

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

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 $\int \sec \theta d \theta =ln| \sec \theta + \tan \theta|+c$

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