$$\int \frac{1}{1+\tan x}dx,$$

A substitution like $t = \tan x, \;dt = (1+t^2)dx$ etc. immediately comes to mind, but I find this method a bit lengthy with the partial fractions. Is there a more concise solution to this?



Multiply the integrand by $\cos x$, we get \begin{equation} \frac{\cos x}{\cos x+\sin x} \end{equation} then let \begin{equation} I=\int\frac{\cos x}{\cos x+\sin x}dx \end{equation} and \begin{equation} J=\int\frac{\sin x}{\cos x+\sin x}dx \end{equation} Find $I+J$ and $I-J$, where $I-J$ can be found by using \begin{equation} \int\frac{f'(x)}{f(x)}dx=\ln\left|f(x)\right|+C \end{equation}



For $\int\dfrac{a\cos x+b\sin x}{c\sin x+d\cos x}dx,$ where at least one of $a,b$ is non-zero

write $a\cos x+b\sin x=A(c\sin x+d\cos x)+B\dfrac{d(c\sin x+d\cos x)}{dx}\ \ \ \ (1)$

So, $\int\dfrac{a\cos x+b\sin x}{c\sin x+d\cos x}dx=A\int\ dx+B\dfrac{d(c\sin x+d\cos x)}{(c\sin x+d\cos x)dx}dx$ $=Ax+\int\dfrac{d(c\sin x+d\cos x)}{(c\sin x+d\cos x)}=Ax+B\ln|c\sin x+d\cos x|+K$

Now from $(1)$, $a\cos x+b\sin x=A(c\sin x+d\cos x)+B(c\cos x-d\sin x)=(Ac-Bd)\sin x+(Ad+Bc)\cos x$

and equating the coefficients of $\cos x,\sin x;$ $Ac-Bd=b,Ad+Bc=a$

This simultaneous equation can be easily solved for $A,B$ in terms of $a,b,c,d$(given)

Can you recognize $a,b,c,d$ here?

  • $\begingroup$ @Hernandez, Hope this is not copy of the other answer $\endgroup$ – lab bhattacharjee Oct 1 '14 at 9:47
  • $\begingroup$ Assuredly, its not, but clarify your method with respect to my question? I'm a tad confused about the 'dx' being a denominator in the general expression. $\endgroup$ – Hernandez Oct 4 '14 at 0:21
  • $\begingroup$ @Hernandez, Please find the edited version $\endgroup$ – lab bhattacharjee Oct 4 '14 at 4:31
  • $\begingroup$ Ah, now I see. Converting the numerator into the sum of a multiple of the denominator and its derivative, for an easy log evaluation. Very, very nice. $\endgroup$ – Hernandez Oct 5 '14 at 4:57

Note that

$$\tan\left(x-{\pi\over4}\right)={\tan x-\tan\left(\pi\over4\right)\over1+\tan x\tan\left(\pi\over4\right)}={\tan x-1\over1+\tan x}=1-{2\over1+\tan x}$$


$$\int{dx\over1+\tan x}={1\over2}\int\left(1-\tan\left(x-{\pi\over4}\right)\right)dx={1\over2}\left(x+\ln\left|\cos\left(x-{\pi\over4}\right)\right| \right)+c\\ ={1\over2}\left(x+\ln|\cos x+\sin x|-\ln\sqrt2 \right)+c\\ ={1\over2}(x+\ln|\cos x+\sin x|)+C$$


$$ I=\int\frac{1}{1+\tan x}dx=\int\frac{\cos x}{\sin x+\cos x}dx=\frac{1}{\sqrt{2}}\int\frac{\cos x}{\sin(x+\frac{\pi}{4})}dx $$ Put $t=x+\frac{\pi}{4}\implies dt=dx$ $$ I=\frac{1}{\sqrt{2}}\int\frac{\cos(t-\frac{\pi}{4})}{\sin t}dt=\frac{1}{2}\int(\cot t+1)dt=\frac{1}{2}\log|\sin t|+\frac{t}{2}+C_1\\ =\frac{x}{2}+\frac{1}{2}\log|\sin(x+\pi/4)|+C_2=\frac{x}{2}+\frac{1}{2}\log|\sin x+\cos x|+C $$


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