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$$\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$$ $$\tan x + \cot x + \csc x + \sec x=\frac{\sin x + 1}{\cos x} +\frac{\cos x + 1}{\sin x} $$ $$= \frac{\sin x +\cos x +1}{\sin x \cos x}$$ $$t= \tan {\frac{x}{2}}$$ 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?

Martin Sleziak
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Aakash Kumar
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    $x=2\arctan(z)$ and everything becomes elementary – tired Jul 26 '16 at 16:16
  • You can multiply and divide by $\csc x \sec x$. After that use your substitution $t = \tan \frac{x}{2}$ will take you through. – Shailesh Jul 26 '16 at 16:18
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    @tired It will increase the denominator and this will be difficult to integrate $$\frac{2z(1-z)}{(1+z^2)^2}{dz}$$ – Aakash Kumar Jul 26 '16 at 16:54
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    @AakashKumar This is asking for partial fractions decomposition. Your denom is already factored. Tedious but doable. Set up: $\frac{Ax+B}{1+z^2}+\frac{Cx+D}{(1+z^2)^2}$ – imranfat Jul 26 '16 at 17:34
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    @AakashKumar where exactly do you see the difficulties? – tired Jul 26 '16 at 17:37
  • @imranfat pfd is not necessary. just notice that $\partial_z\left(\frac{1}{1+z^2}\right)=\frac{2z}{(1+z^2)^2}$ – tired Jul 26 '16 at 17:40
  • @tired I didn't catch that, I straight away went with the "sledge hammer" method. But the point I was trying to make to the OP is that these kind of fractional terms normally ought to be done with PFD. Is he able to do that? And hence your question to him regarding the "difficulties" is valid. – imranfat Jul 26 '16 at 17:41
  • @tired. By the way, what's up with that cat??? – imranfat Jul 26 '16 at 17:43
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    @imranfat it is just tired i guess ;) – tired Jul 26 '16 at 17:46
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    One could also simplify the integrand to $(\cos x+\sin x-1)/2$. Then it is really easy to integrate. – mickep Jul 26 '16 at 19:10
  • @mickep Thanks, it really made it easy. – Aakash Kumar Jul 27 '16 at 03:27

5 Answers5

6

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

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

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

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

user84413
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$$\begin{aligned}\int \frac{1}{\frac{\sin(x)}{\cos(x)}\:+\:\frac{\cos(x)}{\sin(x)}\:+\:\frac{1}{\sin(x)}\:+\:\frac{1}{\cos(x)}}dx & = \int \:\frac{\sin \:\left(2x\right)}{2\left(\cos \:\left(x\right)+\sin \:\left(x\right)+1\right)}dx \\& =\frac{1}{2}\cdot \int \:\frac{\sin \left(2x\right)}{\cos \left(x\right)+\sin \left(x\right)+1}dx \\& =\frac{1}{2}\cdot \frac{1}{2}\cdot \int \:\frac{\sin \left(t\right)}{\sin \left(\frac{t}{2}\right)+\cos \left(\frac{t}{2}\right)+1}dt \\& =\frac{1}{2}\cdot \frac{1}{2}\cdot \int \:\sin \left(\frac{t}{2}\right)+\cos \left(\frac{t}{2}\right)-1dt \\& =\color{red}{\frac{1}{4}\left(-2x+2\sin \left(x\right)-2\cos \left(x\right)\right)+C} \end{aligned}$$

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

Amarildo
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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}

JamesJ
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$$\begin{aligned}\int \frac{d x}{\tan x+\cot x+\csc x+\sec x}&=\int \frac{\cos x \sin x}{1+\cos x+\sin x} d x \\&= -\frac{1}{2} \int \frac{1-(\cos x+\sin x)^2}{1+\cos x+\sin x} d x \\& = -\frac{1}{2} \int(1-\cos x-\sin x) d x\\&= \frac{1}{2}(-x+\sin x-\cos x)+C\end{aligned} $$

Lai
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By substituting $x=y+\dfrac\pi4$ (some more details here),

$$\begin{align*} & \int \frac{dx}{\tan x+\cot x+\csc x+\sec x} \\ &= \frac12 \int \frac{2\cos^2y-1}{\sqrt2\,\cos y+1} \, dy \\ &= \frac12 \int \left(\sqrt2\,\cos y-1\right) \, dy \\ &= \frac1{\sqrt2} \, \sin y - \frac y2 + C \\ &= \frac{\sin x-\cos x-x}2 + C \end{align*}$$

user170231
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