Vtgyjfy

I'm trying to solve the following integral:

$$int frac1{cos x}mathrm dx$$

I know that $int cos xmathrm dx$ = sin x, but I don't know how to proceed with $frac1{cos x}$.

$$intfrac{dx}{cos x}=intfrac{cos x}{1-sin^2x}dx=frac12intcos xleft(frac1{1-sin x}+frac1{1+sin x}right),dx=$$

$$frac12left[-intfrac{d(1-sin x)}{1-sin x}+intfrac{d(1+sin x)}{1+sin x}right]=frac12logfrac{1+sin x}{1-sin x}+C$$

$$int sec(x)dx$$Let $u=tan(frac{x}{2})$
$$int sec(x)dx=2intfrac{du}{1-u^2}$$ Use partial fractions to get $$int sec(x)dx=intfrac{du}{1-u}-intfrac{du}{1+u}=lnleft(frac{1-u}{1+u}right)+C=lnleft(frac{1-tan(frac{x}{2})}{1+tan(frac{x}{2})}right)+C$$

Utilizing the given hint from greelious we get the following

$$I=int frac{mathrm dx}{cos x}=int frac{cos x}{cos^2 x}mathrm dx=int frac{cos x}{1-sin^2 x}mathrm dx$$

Now by the substitution $y=sin x$ we further get

$$I=int frac{cos x}{1-sin^2 x}mathrm dxstackrel{y=sin x}{=}int frac{mathrm dy}{1-y^2}$$

This on can be done via partial fraction decomposition which gives us

$$I=int frac{mathrm dy}{1-y^2}=frac12int left(frac1{1+y}+frac1{1-y}right)mathrm dy=frac12logleft(frac{1+y}{1-y}right)+c$$

And now resubstitute $y=sin(x)$ gives us

$$I=frac12logleft(frac{1+y}{1-y}right)+c=frac12logleft(frac{1+sin x}{1-sin x}right)+c$$

$$therefore~I=intfrac{mathrm dx}{cos x}=frac12logleft(frac{1+sin x}{1-sin x}right)+c$$

Moreover we can show that this solution equals the elegant one given by Rhys Hughes

$$begin{align*}
&color{red}{frac12logleft(frac{1+sin x}{1-sin x}right)}=logleft(sqrt{frac{1+cos left(x-fracpi2right)}{1-cosleft(x-fracpi2right)}}right)=logleft(cotleft(frac{x-fracpi2}{2}right)right)\&=logleft(cotleft(frac x2-fracpi4right)right)
=logleft(frac{1+cotfrac x2}{1-cot frac x2}right)=logleft(frac{left(cosfrac x2+sinfrac x2right)^2}{cos^2 frac x2-sin^2frac x2}right)\&=logleft(frac{1+2sinfrac x2cosfrac x2}{cos(x)}right)=logleft(frac{1+sin x}{cos x}right)=color{red}{log(sec x+tan x)}
end{align*}$$

A method I've seen is:

$$int sec(x)frac{sec x+tan x}{sec x+tan x}dx=intfrac{(sec x+tan x)'}{sec x+tan x}dx=ln|sec x+tan x|+C$$