# integrating trig function

• Jun 9th 2012, 11:34 AM
Tweety
integrating trig function
Can someone please explain to me how the

$\displaystyle \int secx = ln(sec x + tan x)$

am not even sure how to go about it, but $\displaystyle \int sec x = \int \frac{1}{cos x}$ ?

$\displaystyle \int (cosx)^{-1}$ ? And am not sure how to go from here, any help appreciated.
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• Jun 9th 2012, 11:45 AM
Goku
Re: integrating trig function
You could times sec(x) by $\displaystyle \frac{sec(x)+tan(x)}{sec(x)+tan(x)}$... and then solve from there...
• Jun 11th 2012, 01:25 PM
awkward
Re: integrating trig function
Here is another approach.

$\displaystyle \int \frac{1}{\cos x} \;dx = \int \frac{\cos x}{\cos^2 x} \;dx = \int \frac{\cos x}{1 - \sin^2x}\; dx = \int \frac{du}{1-u^2}$

where we made the substitution $\displaystyle u = \sin x$. Now apply partial fractions and finish up.
• Jun 11th 2012, 01:32 PM
Siron
Re: integrating trig function
Another alternative could be to use the substitution $\displaystyle x=\arcsin(t) \Rightarrow dx=\frac{dt}{\sqrt{1-t^2}}$ and using the fact that $\displaystyle \cos[\arcsin(t)]=\sqrt{1-t^2}$ the integral becomes:
$\displaystyle \int \frac{dx}{\cos(x)}= \int \frac{\frac{dt}{\sqrt{1-t^2}}}{\sqrt{1-t^2}}=\int \frac{dt}{1-t^2}=\int \frac{dt}{(1-t)(1+t)} = \ldots$