How do I evaluate ${\Large\int}\cfrac{\mathrm{sin}x+\mathrm{tan}x}{\mathrm{cos}^{2}x}\mathrm{d}x$?

Question

I've got

$$\int\frac{\sin{x}+\tan{x}}{\cos^{2}{x}} \, dx =\int\frac{\sin{x}}{\cos{x}} \, dx + \int\frac{\tan{x}}{\cos^{2}{x}} \, dx \\ =\int \tan{x} \, dx + \int\frac{\sin{x}\times \cos{x}}{\cos{x}}\times \frac{1}{\cos{x}} \, dx \\ =\int\tan{x} \, dx + \int(\sin{x} \sec{x}) \, dx$$

What am I doing wrong?

Answer 1

Bioche's rules say you make the $u=\cos x$ substitution. Your intetgral can be rewritten as $$ \int\frac{\sin x+\tan x}{\cos^2x}\mathrm d\mkern1mu x=\int\biggl(-\frac{1}{u^2}--\frac{1}{u^3}\biggr)\mathrm d\mkern1mu u$$ I think you can finish from here.

Answer 2

Notice, $$\int\frac{\sin x+\tan x}{\cos^2 x}dx$$ $$=\int\frac{\sin x}{\cos^2 x}dx+\int\frac{\tan x}{\cos^2 x}dx$$ $$=\int(\cos x)^{-2}\sin x\ dx+\int\tan x\sec^2 x dx$$ $$=-\int(\cos x)^{-2}\ d(-\cos x)+\int\tan x\ d(\tan x)$$ $$=-\left(\frac{(\cos x)^{-1}}{-1}\right)+\frac{\tan^2 x}{2}+C$$ $$=\color{red}{\sec x+\frac{\tan^2 x}{2}+C}$$