7
$\begingroup$

$$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$$ $$\tan x + \cot x + \csc x + \sec x=\frac{\sin x + 1}{\cos x} +\frac{\cos x + 1}{\sin x} $$ $$= \frac{\sin x +\cos x +1}{\sin x \cos x}$$ $$t= \tan {\frac{x}{2}}$$ On solving , $$\frac{1}{\tan x + \cot x + \csc x + \sec x}=\frac{t(1- t)}{1+ t^2}$$ $$\implies \int \frac{\tan {\frac{x}{2}}(1-\tan {\frac{x}{2}})}{1+\tan^2 {\frac{x}{2}}}{dx}$$

I think, I have made the things more difficult. How can I proceed further? Is there any better substitution for it?

$\endgroup$
  • 2
    $\begingroup$ $x=2\arctan(z)$ and everything becomes elementary $\endgroup$ – tired Jul 26 '16 at 16:16
  • 1
    $\begingroup$ @tired It will increase the denominator and this will be difficult to integrate $$\frac{2z(1-z)}{(1+z^2)^2}{dz}$$ $\endgroup$ – Aakash Kumar Jul 26 '16 at 16:54
  • 1
    $\begingroup$ @AakashKumar This is asking for partial fractions decomposition. Your denom is already factored. Tedious but doable. Set up: $\frac{Ax+B}{1+z^2}+\frac{Cx+D}{(1+z^2)^2}$ $\endgroup$ – imranfat Jul 26 '16 at 17:34
  • 2
    $\begingroup$ @imranfat it is just tired i guess ;) $\endgroup$ – tired Jul 26 '16 at 17:46
  • 3
    $\begingroup$ One could also simplify the integrand to $(\cos x+\sin x-1)/2$. Then it is really easy to integrate. $\endgroup$ – mickep Jul 26 '16 at 19:10
4
$\begingroup$

$\displaystyle\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}=$

$\displaystyle\int\frac{\sin x\cos x}{1+\sin x+\cos x}\,dx=\int\frac{\sin x\cos x}{1+\sin x+\cos x}\cdot\frac{1-(\sin x+\cos x)}{1-(\sin x+\cos x)}\,dx$

$=\displaystyle\int\frac{\sin x\cos x(1-(\sin x+\cos x))}{1-(\sin x+\cos x)^2}\,dx=\int\frac{\sin x\cos x-\sin^2 x\cos x-\cos^2x\sin x}{-2\sin x\cos x}\,dx$

$\displaystyle=-\frac{1}{2}\int(1-\sin x-\cos x)\,dx=\frac{1}{2}(-x-\cos x+\sin x)+C$

$\endgroup$
2
$\begingroup$

$$\begin{aligned}\int \frac{1}{\frac{\sin(x)}{\cos(x)}\:+\:\frac{\cos(x)}{\sin(x)}\:+\:\frac{1}{\sin(x)}\:+\:\frac{1}{\cos(x)}}dx & = \int \:\frac{\sin \:\left(2x\right)}{2\left(\cos \:\left(x\right)+\sin \:\left(x\right)+1\right)}dx \\& =\frac{1}{2}\cdot \int \:\frac{\sin \left(2x\right)}{\cos \left(x\right)+\sin \left(x\right)+1}dx \\& =\frac{1}{2}\cdot \frac{1}{2}\cdot \int \:\frac{\sin \left(t\right)}{\sin \left(\frac{t}{2}\right)+\cos \left(\frac{t}{2}\right)+1}dt \\& =\frac{1}{2}\cdot \frac{1}{2}\cdot \int \:\sin \left(\frac{t}{2}\right)+\cos \left(\frac{t}{2}\right)-1dt \\& =\color{red}{\frac{1}{4}\left(-2x+2\sin \left(x\right)-2\cos \left(x\right)\right)+C} \end{aligned}$$

Applyied substitution: $$\color{blue}{t=2x,\quad \:dt=2dx}$$

$\endgroup$
0
$\begingroup$

A trigonometric formula can be used: \begin{align} &(\sin x+\cos x+1)(\sin x+\cos x-1)=\sin 2x\\ I&=\int\frac{\sin x\cos x}{\sin x+\cos x+1}dx=\int\frac{1}{2}(\sin x+\cos x-1)dx=\frac{1}{2}(-\cos x+\sin x-x)+C\\ \end{align}

$\endgroup$

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.