Question

pineapplesinla....integral sinx +secx/tanx dx help

Answer 1

\[\int\sin(x)+\frac{\sec(x)}{\tan(x)} dx\]
\[\frac{\sec(x)}{\tan(x)}=\frac{1}{\cos(x)}\times \frac{\cos(x)}{\sin(x)}=\csc(x)\] is a start

Answer 2

I am assuming you intend the following:\[\int\left( \sin x + \dfrac{\sec x}{\tan x}\right) dx\]Solve as follows:\[\int \sin x dx + \int \dfrac{\sec x}{\tan x}dx\]\[-\cos x + \int \dfrac{\left(\dfrac{1}{\cos x}\right)}{\left(\dfrac{\sin x}{\cos x}\right)}dx\]\[-\cos x + \int \dfrac{dx}{\sin x}\]\[-\cos x + \int \csc x dx\]\[-\cos x + \int \csc x\left(\dfrac{
\csc x + \cot x}{\csc x + \cot x} \right)dx\]\[-\cos x + \int \dfrac{
\csc^2 x + \csc x\cot x}{\csc x + \cot x} dx\]\[-\cos x - \int \dfrac{
-\csc^2 x - \csc x\cot x}{\csc x + \cot x} dx\]\[-\cos x - \ln \left|\csc x + \cot x\right|+C\]