2
$\begingroup$

I got this problem in my homework exercise:

$$\int \arctan(\sec x + \tan x) dx$$

I simplified it to

$$\int \arctan\left(\dfrac{1+\sin x}{\cos x}\right) dx$$

$$=\int \arctan\left(\sqrt{\dfrac{1+\sin x}{1-\sin x}}\right) dx$$

now, I tried putting $\sqrt{\dfrac{1+\sin x}{1-\sin x}} = \tan t$

Then $x = \arcsin(\sin^2 t-\cos^2 t)$ but it becomes complete mess after that!

$\endgroup$
13
$\begingroup$

$${1+\sin(x)\over \cos(x)}={\sin(x/2)+\cos(x/2)\over \cos(x/2)-\sin(x/2)}={1+\tan(x/2)\over 1-\tan(x/2)}=\tan(\pi/4+x/2)$$

Also keep in mind: $$\tan^{-1}(\tan(z))=\begin{cases}z&-\pi/2\le z\le\pi/2\\ z-\pi&\ \ \ \pi/2\lt z\le \pi\\z+\pi&\ \ -\pi\le z\lt-\pi/2 \end{cases}$$

$\endgroup$
  • $\begingroup$ That was fast. Did you do this before? $\endgroup$ – N.S.JOHN Aug 15 '16 at 7:15
  • 2
    $\begingroup$ @N.S.JOHN Had a lot of practice in trigonometry and integration in 11th and 12th standard ;-) $\endgroup$ – Qwerty Aug 15 '16 at 7:16
  • $\begingroup$ @N.S.JOHN Since $\arctan(\tan(x))$ is not always equal to $x$, the primitive you are looking for is not $\int (\pi/4+x/2)\,dx$. $\endgroup$ – Robert Z Aug 15 '16 at 7:36
  • 2
    $\begingroup$ I think My edit will help @N.S.JOHN about the problem he was about to face. $\endgroup$ – Qwerty Aug 15 '16 at 7:51
5
$\begingroup$

An direct way to use the integration by parts:

Let $u = \arctan(\sec(x)+\tan(x))$ then $du = \frac{1}{2} dx$

$dv = dx$ then $v = x$.

Then the integral becomes $$ x \arctan(\sec(x)+\tan(x)) - \int \frac{x}{2}dx$$

$\endgroup$
4
$\begingroup$

Using integration by parts, we get:

$$\int \arctan(\sec x + \tan x) dx=x\arctan(\sec x + \tan x)-\dfrac{x^2}{4}+C$$

$\endgroup$
2
$\begingroup$

By integration by parts we obtain $$\int \arctan(\sec x + \tan x) dx=x\arctan(\sec x + \tan x)-\dfrac{x^2}{4}+C.$$ Since we have the identity (see Qwerty's answer): $$\sec x + \tan x={1+\sin(x)\over \cos(x)}=\tan(\pi/4+x/2),$$ the primitive can be simplified (!?) to $$x\arctan(\sec x + \tan x)-\dfrac{x^2}{4}+C=x\arctan(\tan(\pi/4+x/2))-\dfrac{x^2}{4}+C\\=\frac{x^2+\pi x}{4}-\pi x \left\lfloor \frac{x}{2\pi}+\frac{3}{4}\right\rfloor+C.$$

P.S. Note that if we erroneously write that $\arctan(\sec x + \tan x)=\pi/4+x/2$, then we get a different answer $$\int \arctan(\sec x + \tan x) dx=\int (\pi/4+x/2) dx=\frac{x^2+\pi x}{4}+C.$$

$\endgroup$
1
$\begingroup$

sec x + tan x = tan t. then sec x - tan x = cot t. sec x = (cot t + tan t)/2. (sec t)^2 dt = sec x(sec x + tan x)dx = (tan t(tan t + cot t)/2)dx. Substitute dx and simplify.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.