Question

Solve $\int {\dfrac{{(1 - \cot x)}}{{(1 + \cot x)}}} dx$

Answer 1

Hint : To solve this question first convert the $\cot x$ trigonometric function to $\sin x$ and $\cos x$. Then simplify the equation and then take the denominator to another variable and try to adjust the upper equation in the derivative of that variable. Then integrate that function and again substitute the value of that variable in order to find the integration in the original variable.

Complete step-by-step answer :
Let the given integration is equal to $I = \int {\dfrac{{(1 - \cot x)}}{{(1 + \cot x)}}} dx$
Now on converting that equation in $\sin x$ and $\cos x$ function.
$I = \int {\dfrac{{(1 - \dfrac{{\sin x}}{{\cos x}})}}{{(1 + \dfrac{{\sin x}}{{\cos x}})}}} dx$
Now taking LCM
$I = \int {\dfrac{{(\dfrac{{\cos x - \sin x}}{{\cos x}})}}{{(\dfrac{{\cos x + \sin x}}{{\cos x}})}}} dx$
On further solving this equation
$I = \int {\dfrac{{(\cos x - \sin x)}}{{(\sin x + \cos x)}}} dx$
Let, $\sin x + \cos x = t$
Differentiating the equation both side
$(\cos x - \sin x)dx = dt$
This term is the nominator of the integration
$I = \int {\dfrac{{dt}}{t}}$
Integration of $\int {\dfrac{{dt}}{t} = \ln t}$
$I = \ln t + c$
Now again put the value of t in the last equation
$I = \ln (\sin x + \cos x) + c$
Here, c is the constant term that comes with the integration
The integration of the equation: $I = \int {\dfrac{{(1 - \cot x)}}{{(1 + \cot x)}}} dx = \ln (\sin x + \cos x) + c$

Note : Integration is usually used to find the area of the curves whose equations are given. If the sign is mentioned with the value of integration then that doesn’t mean that area is negative but that is indicating the area that given area is on the side of the positive axis or on the side of the negative axis.