Definite Integral (Trig Substitution) with answer containing sine and arctan?

[eric_f]

I've been racking my brain for a while on this one and can't seem to come by a cleaner answer....Is there something obvious I'm missing? See attached.

 

[eric_f]

I forgot that 1/sin = cosecant. Came to my senses ths morning and figured it out...

 

[HallsofIvy]

I would think it would be easier to recall that $\displaystyle tan(\theta)= \frac{4}{3}$ for a right triangle with legs of length 4 and 3. And then the hypotenuse has length 5. So $\displaystyle sin(arctan(4/3))= \frac{4}{5}$ and so $\displaystyle \frac{1}{sin(arctan(4/3)}= \frac{5}{4}$. Similarly, $\displaystyle tan(\theta)= 2/3$ for a right triangle with legs of length 2 and 3 and so hypotenuse $\displaystyle \sqrt{13}$. $\displaystyle sin(arctan(2/3))= \frac{2}{\sqrt{13}}$ and, of course, $\displaystyle \frac{1}{sin(arctan(2/3))}= \frac{\sqrt{13}}{2}$.

 

[DrPhil]

I've been racking my brain for a while on this one and can't seem to come by a cleaner answer....Is there something obvious I'm missing? See attached.

View attachment 3389

As a first step, I got rid of the numbers in the radical by letting $\displaystyle a=3/2$ and pulling 1/2 outside the integral:

$\displaystyle \displaystyle \dfrac 1 2 \int_1^2\dfrac{dx}{x^2\ \sqrt{x^2 + a^2}}$

When I don't recognize what to do with an integrand, I look it up in a Table of Integrals. That gives me clues about substitutions, integration by parts, etc.. In this case I found the antiderivative to be a relatively simple function:

If $\displaystyle r = \sqrt{x^2 + a^2}$, then $\displaystyle \displaystyle \int\dfrac{dx}{x^2\ r} = -\dfrac{r}{a^2\ x}$

Does that give you any guidance in your substitutions?