integrating trig function

**Tweety** · June 9th 2012, 11:34 AM

Can someone please explain to me how the

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

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

$\int (cosx)^{-1}$ ? And am not sure how to go from here, any help appreciated.
.

**Goku** · June 9th 2012, 11:45 AM

You could times sec(x) by $\frac{sec(x)+tan(x)}{sec(x)+tan(x)}$ ... and then solve from there...

**awkward** · June 11th 2012, 01:25 PM

Here is another approach.

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

where we made the substitution $u = \sin x$ . Now apply partial fractions and finish up.

**Siron** · June 11th 2012, 01:32 PM

Another alternative could be to use the substitution $x=\arcsin(t) \Rightarrow dx=\frac{dt}{\sqrt{1-t^2}}$ and using the fact that $\cos[\arcsin(t)]=\sqrt{1-t^2}$ the integral becomes:
$\int \frac{dx}{\cos(x)}= \int \frac{\frac{dt}{\sqrt{1-t^2}}}{\sqrt{1-t^2}}=\int \frac{dt}{1-t^2}=\int \frac{dt}{(1-t)(1+t)} = \ldots$

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