**Question:**

Evaluate the integral

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

**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, $I=\int \frac{1}{1-\cot x} \mathrm{dx}$

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, $I=\int \frac{1}{1-\cot x} d x=\int \frac{1}{1-\frac{\cos x}{\sin x}} d x=\int \frac{\sin x}{\sin x-\cos x} d x$

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

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

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

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

Comparing both sides we have:

$C=0$

$A-B=0 \Rightarrow A=B$

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

$\therefore A=B=1 / 2$

Thus I can be expressed as:

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

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

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

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

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

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

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

$I_{1}=\frac{1}{2} \int \frac{d u}{u}=\frac{1}{2} \log |u|+C_{1}$

$\therefore I_{1}=\frac{1}{2} \log |\sin x-\cos x|+C_{1} \ldots \ldots$ equation 2

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

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

From equation 1,2 and 3 we have:

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

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

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