# Thread: Derivatives and Integrals with logarithmic functions

1. ## Derivatives and Integrals with logarithmic functions

We went over this stuff in class, and I was fine doing the examples, but now I can't do the homework.
The first is:
Integral of (sec(x)dx)/(sqrt(secx+tanx))
Any hints or ideas of where I can start?

Also:

Find the derivative, logarithmically, of y = (x*sin(x))/sqrt(sec(x))
I tried this one just using the quotient rule for derivatives, and it turned out pretty bad.

2. Originally Posted by mistykz
Find the derivative, logarithmically, of y = (x*sin(x))/sqrt(sec(x))
I tried this one just using the quotient rule for derivatives, and it turned out pretty bad.
Take the hint:

lny=lnx+ln(sinx)+0.5ln(cosx)

and differentiate.....

3. Hello, mistykz!

What you covered in class may not have prepared you for this integral.

$\int \frac{\sec x\,dx}{\sqrt{\sec x + \tan x}}$
Multiply by $\frac{\sec x + \tan x}{\sec x + \tan x}$

$\int \frac{\sec x}{(\sec x + \tan x)^{\frac{1}{2}}} \cdot\frac{\sec x + \tan x}{\sec x + \tan x}\cdot dx \;=\;\int\frac{\sec x\tan x + \sec^2\!x}{(\sec x + \tan x)^{\frac{3}{2}}} \,dx$

Now let $u \:=\:\sec x + \tan x$

Use logarithmic differentiation: . $y \:= \:\frac{x\sin x}{\sqrt{\sec x}}$
If you differentiated logarithmically, there are no quotients involved!

Take logs: . $\ln y \;=\;\ln\left(\frac{x\sin x}{(\sec x)^{\frac{1}{2}}}\right) \;=\;\ln(x) + \ln(\sin x) - \frac{1}{2}\ln(\sec x)$

Differentiate implicitly: . $\frac{1}{y}\cdot y' \;=\;\frac{1}{x} + \frac{\cos x}{\sin x} -\frac{1}{2}\frac{\sec x\tan x}{\sec x} \;=\;\frac{1}{x} + \cot x - \frac{1}{2}\tan x$

Then: . $y' \;=\;y\left(\frac{1}{x} + \cot x - \frac{1}{2}\tan x\right) \;=\;\frac{x\sin x}{\sqrt{\sec x}}\left(\frac{1}{x} + \cot x - \frac{1}{2}\tan x\right)$