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

Solution 1:

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!

Solution 2:

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 $$ \int \frac{dx}{x + \sqrt{x^2 + 16} } = \log \left( x + \sqrt{x^2 + 16} \right) + C_2 \; . $$