How do I integrate $\sec(x)$?

Question

My HW asks me to integrate $\sin(x)$, $\cos(x)$, $\tan(x)$, but when I get to $\sec(x)$, I'm stuck.

Answer 1

Hint: I'm assuming you know the derivatives of $\sec(x)$ and $\tan(x)$.

Since $\frac d{dx}\tan(x)=\sec^2(x)$ and $\frac d{dx}\sec(x)=\sec(x)\tan(x)$,

therefore $\frac d{dx}[\tan(x)+\sec(x)]=\sec(x)[\tan(x)+\sec(x)]$

Note that here: $\sec(x)=\dfrac{\frac d{dx}[\tan(x)+\sec(x)]}{\tan(x)+\sec(x)}$

Now let $u=\tan(x)+\sec(x)$, you get $\sec(x)=\dfrac{du}u$, then you integrate both sides:

\begin{align*}\therefore \int\sec(x)\ dx&=\int\frac{du}u\\\\&=\ln|u|+C\\\\&=\ln|\tan(x)+\sec(x)|+C\end{align*}

Answer 2

$$\int \frac{1}{\cos x} dx =\int \frac{\cos x}{\cos^2 x} dx =\int \frac{d(\sin x)}{1-\sin^2 x}$$ $$=\frac {1}{2}\int (\frac {1}{1-\sin x}+\frac {1}{1+\sin x}) d(\sin x) =\frac {1}{2}\int \frac {d(\sin x)}{1-\sin x}+\frac {1}{2}\int \frac {d(\sin x)}{1+\sin x} $$ $$= -\frac {1}{2}\ln |1-\sin x|+\frac {1}{2}\ln |1+\sin x|+C$$ $$=\frac {1}{2}\ln|\frac {1+\sin x}{1-\sin x}|+C=\frac {1}{2}\ln|\frac {(1+\sin x)^2}{\cos^2 x}|+C$$ $$=\ln|\sec x+\tan x|+C$$ Hope it can help you

Answer 3

$$\int \sec x dx$$

Using Weierstrass substitution,

$$\tan x=\tan(\frac{x}{2}+\frac{x}{2}) \iff \tan(\frac{x}{2}+\frac{x}{2})=\frac{2\tan\frac{x}{2}}{1-\tan^2(\frac{x}{2})}$$

$$ \tan(\frac{x}{2}+\frac{x}{2})=\frac{2\tan\frac{x}{2}}{1-\tan^2(\frac{x}{2})}\iff \tan x= \frac{2t}{1-t^2}$$

$$\tan(\frac{x}{2})=t \iff \frac{1}{2}\sec^2(\frac{x}{2})dx=dt$$

$$\frac{1}{2}\sec^2(\frac{x}{2})dx=dt \iff dx= 2\cos^2\frac{x}{2}dt$$

$$dx= 2\cos^2\frac{x}{2}dt \iff dx=\frac{2}{(1+t^2)}dt$$

$$\cos(x)=\cos(\frac{x}{2}+\frac{x}{2}) \iff \cos( \frac{x}{2}+ \frac{x}{2})= \cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})$$

$$\cos^2(\frac{x}{2})-\sin^2(\frac{x}{2})=\frac{1}{t^2+1}-\frac{t^2}{t^2+1} \iff \frac{1-t^2}{t^2+1}=\cos x$$

$$\frac{1-t^2}{t^2+1}=\cos x \iff \sec x=\frac{1+t^2}{1-t^2} $$

$$\int \sec x dx=\int \frac{1+t^2}{1-t^2}. \frac{2}{(1+t^2)}dt \iff \int \frac{1+t^2}{1-t^2}. \frac{2}{(1+t^2)}dt=\int \frac{2}{1-t^2}dt$$

Partial fractions!

$$\frac{2}{1-t^2}\equiv \frac{A}{1-t}+\frac{B}{1+t}\iff 2\equiv A(1+t)+B(1-t)$$

Solution is $t=1, A=1,t=-1,B=1$

$$\int \frac{1}{1+t}+\frac{1}{1-t}dt=\ln|1+t|-\ln|1-t|+c \iff \ln|\frac{1+t}{1-t}|.\ln|\frac{1+t}{1+t}|+c=\ln|\frac{(1+t)^2}{1-t^2}|+c$$

$$\ln|\frac{(1+t)^2}{1-t^2}|+c=\ln|\frac{t^2+2t+1}{1-t^2}|+c \iff \ln|\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2}|$$

So,

$$\ln|\frac{1+t^2}{1-t^2}+\frac{2t}{1-t^2}| \iff \ln|\sec x + \tan x|$$

This is a standard integral for $\sec x$

Conclusion!

$$\int \sec xdx=\ln|\sec x+\tan x|+c$$