# Thread: Solve The Following Trigonometric Integral.

1. ## Solve The Following Trigonometric Integral. (1)

Hi
i tried to do it many times
but really its hard
its from an incomplete exam

$\displaystyle \int \frac{1-tan(x)}{1+tan(x)}dx$

2. Originally Posted by TWiX
Hi
i tried to do it many times
but really its hard
its from an incomplete exam

$\displaystyle \int \frac{1-tan(x)}{1+tan(x)}dx$
Hello .
Multiply the integrated function by $\displaystyle \frac{1-tan(x)}{1-tan(x)}$
You will get:

$\displaystyle \int\frac{(1-tan(x))^2}{1-tan^2(x)}dx$
=$\displaystyle \int\frac{tan^2(x)-2tan(x)+1}{1-tan^2(x)}dx$
$\displaystyle =\int\frac{sec^2(x)-2tan(x)}{1-tan^2(x)}dx$
Multiply the integrated function by $\displaystyle \frac{cos^2(x)}{cos^2(x)}$, And with some trigonometric identities, you will face:
$\displaystyle \int\frac{1-sin(2x)}{cos(2x)}dx$

Now, It looks easier. Right ?

3. See attachment

4. Hello, =TWiX!

$\displaystyle \int \frac{1-\tan x}{1+\tan x}\,dx$
The function is: .$\displaystyle \frac{1-\tan x}{1 + \tan x} \;=\;\frac{1-\frac{\sin x}{\cos x}}{1 + \frac{\sin x}{\cos x}}$

Multiply by $\displaystyle \frac{\cos x}{\cos x}\!:\quad\frac{\cos x\left(1 - \frac{\sin x}{\cos x}\right)}{\cos x\left(1 + \frac{\sin x}{\cos x}\right)} \;=\; \frac {\cos x - \sin x}{\cos x + \sin x}$

The integral becomes: .$\displaystyle \int\frac{\cos x - \sin x}{\cos x + \sin x}\,dx$

Let: .$\displaystyle u \:=\:\cos x + \sin x \quad\Rightarrow\quad du \:=\:(-\sin x + \cos x)\,dx \quad\Rightarrow\quad du \:=\:(\cos x - \sin x)\,dx$

Substitute: .$\displaystyle \int\frac{du}{u}\quad\hdots$ Got it?

5. Thanks all.