Vrftsjtryk

$$begin{align}
int sec x , dx
&= int cos x left( frac{1}{cos^2x} right) , dx \
&= int cos x left( frac{1}{1-sin^2x} right) , dx \
& = intcos xcdotfrac{1}{1-frac{1-cos2x}{2}} , dx \
&= int cos x cdotfrac{2}{1+cos2x} , dx
end{align}$$

I am stuck in here. Any help to integrate secant?

begin{align*}intsec x,mathrm dx&=intfrac1{cos x},mathrm dx\&=intfrac{cos x}{cos^2x},mathrm dx\&=intfrac{cos x}{1-sin^2x},mathrm dx.end{align*}
Now, doing $sin x=t$ and $cos x,mathrm dx=mathrm dt$, you get $displaystyleintfrac{mathrm dt}{1-t^2}$. Butbegin{align*}intfrac{mathrm dt}{1-t^2}&=frac12intfrac1{1-t}+frac1{1+t},mathrm dt\&=frac12left(-log|1-t|+log|1+t|right)\&=frac12logleft|frac{1+t}{1-t}right|\&=frac12logleft|frac{(1+t)^2}{1-t^2}right|\&=logleft|frac{1+t}{sqrt{1-t^2}}right|\&=logleft|frac{1+sin x}{sqrt{1-sin^2x}}right|\&=logleft|frac1{cos x}+frac{sin x}{cos x}right|\&=log|sec x+tan x|.end{align*}

An alternative method: The trick here is to multiply $sec{x}$ by $dfrac{tan{x}+sec{x}}{tan{x}+sec{x}}$, then substitute $u=tan{x}+sec{x}$ and $du=(sec^2{x}+tan{x}sec{x})~dx$:

$$int sec{x}~dx=int sec{x}cdot frac{tan{x}+sec{x}}{tan{x}+sec{x}}~dx=int frac{sec{x}tan{x}+sec^2{x}}{tan{x}+sec{x}}~dx=int frac{1}{u}~du=cdots$$

Not obvious, though it is efficient.

After $int cos x left(frac{1}{1-sin^2x}right)dx$ use the transformation $z = sin x$ and $dz = cos x , dx$.

Edit:

$$intfrac{1}{1-u^2},du = frac{1}{2}intfrac{(1+u)+(1-u)}{(1+u)(1-u)} = frac{1}{2} int frac{1}{1+u} + frac{1}{1-u},du$$

And use, $int frac{1}{u},du = ln|u|$

Although the integral can be evaluated in a straightforward way using real analysis, I thought it might be instructive to present an approach based on complex analysis. To that end, we now proceed.

We use Euler's Formula, $e^{ix}=cos(x)+isin(x)$, to write $displaystyle sec(x)=frac2{e^{ix}+e^{-ix}}=frac{2e^{ix}}{1+e^{i2x}}$. Then, we have

$$begin{align}
int sec(x),dx&=int frac2{e^{ix}+e^{-ix}}\\
&=int frac{2e^{ix}}{1+e^{i2x}},dx \\
&=-i2 int frac{1}{1+(e^{ix})^2},d(e^{ix})\\
&=-i2 arctan(e^{ix})+Ctag 1\\
&=logleft(frac{1-ie^{ix}}{1+ie^{ix}}right)+Ctag2\\
&=logleft(-ileft(frac{1+sin(x)}{icos(x)}right)right)+Ctag3\\
&=log(sec(x)+tan(x))+C'tag4
end{align}$$

NOTES:

In going from $(1)$ to $(2)$, we used the identity $arctan(z)=i2logleft(frac{1-iz}{1+iz}right) $

In going from $(2)$ to $(3)$, we multiplied the numerator and denominator of the argument of the logarithm function by $1-ie^{ix}$. Then, we used

$$frac{1-ie^{ix}}{1+ie^{ix}}=frac{-i2cos(x)}{2(1-sin(x))}=-ifrac{1+sin(x)}{cos(x)}$$

Finally, in going from $(3)$ to $(4)$, we absorbed the term $log(-i)$ into the integration constant $C$ and labeled the new integration constant $C'=C+log(-i)$.

Just to spell out Lord Shark the Unknown's suggestion, $$t=tanfrac{x}{2}impliessec x=frac{1+t^2}{1-t^2},,dx=frac{2dt}{1+t^2}impliesintsec xdx=intfrac{2 dt}{1-t^2}.$$From that point on, the same partial-fractions treatment as in multiple other answers can be used. Admittedly the expression thus obtained for the antiderivative is $lnleft|frac{1+tanfrac{x}{2}}{1-tanfrac{x}{2}}right|+C$ instead of $lnleft|frac{1+sin x}{1-sin x}right|+C$ or $ln|sec x+tan x|+C$, but of course they're all the same thanks to suitable trigonometric identities (again, obtainable by writing things as functions of $tanfrac{x}{2}$).