simplifying sin(arctan[x/2])

[ahorn]

Hi

I just differentiated $\displaystyle \int\frac{1}{x^2\sqrt{x^2+4}}dx$ by substituting $\displaystyle x=2\tan\theta$. I ended up with $\displaystyle -\frac{1}{\sin\left(\arctan\left(\frac{x}{2}\right)\right)}+C$. The solution given in the textbook is $\displaystyle -\frac{\sqrt{4-x^2}}{4x}+C$. How do I simplify my answer?

Thank you in advance.

 

[HallsofIvy]

Here is a simple way: imagine a right triangle with one angle $\displaystyle \theta$, the "opposite side" of length x/2 and "near side" of length 1. Tangent= "opposite side over near side": $\displaystyle tan(\theta)= (x/2)/1= x/2$ so that $\displaystyle \theta= arctan(x/2)$.

By the Pythagorean theorem, the hypotenuse of that right triangle has length $\displaystyle \sqrt{(x/2)^2+ 1}= \sqrt{x^2/4+ 1}= \frac{\sqrt{x^2+ 4}}{2}$.

$\displaystyle sin(arctan(x/2))= sin(\theta)$ equals "opposite side over hypotenuse" $\displaystyle = \frac{x}{2}\frac{2}{\sqrt{x^2+ 4}}= \frac{x}{\sqrt{x^2+ 4}}$.

 

[Deleted member 4993]

Hi

I just differentiated $\displaystyle \int\frac{1}{x^2\sqrt{x^2+4}}dx$ by substituting $\displaystyle x=2\tan\theta$. I ended up with $\displaystyle -\frac{1}{\sin\left(\arctan\left(\frac{x}{2}\right)\right)}+C$. The solution given in the textbook is $\displaystyle -\frac{\sqrt{4-x^2}}{4x}+C$. How do I simplify my answer?

Thank you in advance.

Did you check your answer - by differentiating it. I believe you are missing a "multiplicative constant".

 

[Quaid]

I just differentiated $\displaystyle \int\frac{1}{x^2\sqrt{x^2+4}}dx$

You meant to write, "I just integrated...", correct?