I'm having problems integrating
$\displaystyle
\int\frac {sin(x)}{cos^2(x)}dx
$
the answer in the book is
$\displaystyle
\frac{1}{cos(x)}+c
$
I can't see how this answer is obtained.
Hello, macca101!
Another method . . .
$\displaystyle \int\frac{\sin(x)}{\cos^2(x)}\,dx$
The answer in the book is: .$\displaystyle \frac{1}{\cos(x)}+C
$
I'm surprised that they left the answer like that . . .
We have: .$\displaystyle \int\frac{\sin x}{\cos^2x}\,dx\;=\;\int\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}\,dx\;=$ $\displaystyle \int\sec x\tan x\,dx \;= \;\sec x + C$