Is there anyone who could walk me through this practice problem?:
$\displaystyle \int{\sqrt{1 + \frac{cosx}{sinx}}}\,dx$
I really struggle with integrals involving radicals or trigonometry.
Hey infraRed.
There is probably a simpler way but one suggestion I have is to use complex numbers. cos(x) = [e^(ix) + e^(-ix)]/2 and sin(x) = [e^(ix) - e^(-ix)]/2i.
Given that you probably haven't done a lot of stuff like that can you mention the identities and concepts that you have studied in the course that you are taking?
Hello, infraRed!
$\displaystyle \displaystyle \int \sqrt{1 + \left(\frac{\cos x}{\sin x}\right)^2}\,dx $
Apply some trig identities . . .
$\displaystyle \sqrt{1 + \left(\frac{\cos x}{\sin x}\right)^2} \;=\; \sqrt{1 + \cot^2x} \;=\;\sqrt{\csc^2x} \;=\;\csc x$
So we have: .$\displaystyle \displaystyle \int \csc x\,dx$
Got it?
Which of course then can be written as
$\displaystyle \displaystyle \begin{align*} \int{\csc{(x)}\,dx} &= \int{\frac{1}{\sin{(x)}}\,dx} \\ &= \int{ \frac{\sin{(x)}}{\sin^2{(x)}}\,dx} \\ &= -\int{\frac{-\sin{(x)}}{1 - \cos^2{(x)}}\,dx} \\ &= -\int{\frac{1}{1 - u^2}\,du} \textrm{ after making the substitution } u = \cos{(x)} \implies du = -\sin{(x)}\,dx \end{align*}$
I'm sure you can go from here