# Evaluate the integral

Question:

Evaluate the integral

$\int \frac{1}{1-\tan x} d x$

Solution:

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos, polynomial, log etc and formula for some

special functions.

Let, $I=\int \frac{1}{1-\tan x} d x$

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form $\int \frac{a \sin x+b \cos x+c}{d \sin x+e \cos x+f} d x$

Then substitute numerator as -

$a \sin x+b \cos x+c=A \frac{d}{d x}(d \sin x+e \cos x+f)+B(d \sin x+e \cos x+c)+C$

Where A, B and C are constants

We have, $I=\int \frac{1}{1-\tan x} d x=\int \frac{1}{1-\frac{\sin x}{\cos x}} d x=\int \frac{\cos x}{\cos x-\sin x} d x$

As I matches with the form described above, So we will take the steps as described.

$\therefore \cos x=A \frac{d}{d x}(\cos x-\sin x)+B(\cos x-\sin x)+C$

$\Rightarrow \cos x=A(-\sin x-\cos x)+B(\cos x-\sin x)+C\left\{\because \frac{d}{d x} \cos x=-\sin x\right\}$

$\Rightarrow \cos x=-\sin x(B+A)+\cos x(B-A)+C$

Comparing both sides we have:

$C=0$

$B-A=1 \Rightarrow A=B-1$

$B+A=0 \Rightarrow 2 B-1=0 \Rightarrow B=1 / 2$

$\therefore A=B-1=-1 / 2$

Thus I can be expressed as:

$I=\int \frac{\frac{1}{2}(\cos x+\sin x)+\frac{1}{2}(\cos x-\sin x)}{(\cos x-\sin x)} d x$

$I=\int \frac{\frac{1}{2}(\cos x+\sin x)}{(\cos x-\sin x)} d x+\int \frac{\frac{1}{2}(\cos x-\sin x)}{(\cos x-\sin x)} d x$

$\therefore$ Let $I_{1}=\frac{1}{2} \int \frac{(\cos x+\sin x)}{(\cos x-\sin x)} d x$ and $I_{2}=\frac{1}{2} \int \frac{(\cos x-\sin x)}{(\cos x-\sin x)} d x$

$\Rightarrow I=I_{1}+I_{2} \ldots$ equation 1

$I_{1}=\frac{1}{2} \int \frac{(\cos x+\sin x)}{(\cos x-\sin x)} d x$

Let, $u=\cos x-\sin x \Rightarrow d u=-(\cos x+\sin x) d x$

So, I $_{1}$ reduces to:

$\mathrm{I}_{1}=-\frac{1}{2} \int \frac{\mathrm{du}}{\mathrm{u}}=-\frac{1}{2} \log |\mathrm{u}|+\mathrm{C}_{1}$

$\therefore \mathrm{I}_{1}=-\frac{1}{2} \log |\cos \mathrm{x}-\sin \mathrm{x}|+\mathrm{C}_{1} \ldots . .$ equation 2

As, $I_{2}=\frac{1}{2} \int \frac{(\cos x-\sin x)}{(\cos x-\sin x)} d x=\frac{1}{2} \int d x$

$\therefore I_{2}=\frac{x}{2}+C_{2} \ldots . .$ equation 3

From equation 1,2 and 3 we have:

$\mathrm{I}=-\frac{1}{2} \log |\cos \mathrm{x}-\sin \mathrm{x}|+\mathrm{C}_{1}+\frac{\mathrm{x}}{2}+\mathrm{C}_{2}$

$\therefore I=-\frac{1}{2} \log |\cos x-\sin x|+\frac{x}{2}+C$