Question

Integrate $\int {\dfrac{1}{{1 + \tan x}}} dx$

Answer 1

Hint : The simple meaning of trigonometry is calculations of triangles.
As we know that, $\tan x = \dfrac{{\sin x}}{{\cos x}}$
By using this identity we will solve this problem. By substituting this value in the given equation, we can calculate the integration as we get terms of the integral in terms of sin and cos.

Complete step-by-step answer :
$\int {\dfrac{1}{{1 + \tan x}}} dx$ = $\int {\dfrac{1}{{1 + \dfrac{{\sin x}}{{\cos x}}}}}$ $dx$
B taking LCM
= $\int {\dfrac{1}{{\dfrac{{\sin x + \cos x}}{{\cos x}}}}}$ $dx$

= $\int {\dfrac{{\cos x}}{{\sin x + cox}}}$ $dx$
Multiply and divide by 2
= $\dfrac{1}{2}\int {\dfrac{{2\cos x}}{{\sin x + \cos x}}}$ $dx$
= $\dfrac{1}{2}\int {\left( {\dfrac{{2\cos x}}{{\sin x + \cos x}} - 1 + 1} \right)}$ $dx$
$\dfrac{1}{2}\int {\dfrac{{2\cos x}}{{\sin x + \cos x}}} - 1 + 1$ $dx$
= $\dfrac{1}{2}\int {\left( {\dfrac{{\cos x + \cos x}}{{\sin x + \cos x}} - \dfrac{{\sin x + \cos x}}{{\sin + \cos x}} + 1} \right)} dx$
= $\dfrac{1}{2}\int {\dfrac{{\cos x - \sin x}}{{\sin x + \cos x}}} dx + \dfrac{1}{2}\int {dx}$
Put , $\sin x + \cos x = u$
Now, differentiate this above equation, we get-
$du = \cos x - \sin x$
By substituting, we will get equation like-
= $\dfrac{1}{2}\int {udu + \dfrac{1}{2}} x$
= $\dfrac{1}{2}\log u + \dfrac{1}{2}x$
By Substituting the value of u, we get-
= $\dfrac{1}{2}\log \left( {\sin x + \cos x} \right) + \dfrac{1}{2}x + c$
$\int {\dfrac{1}{{1 + \tan x}}} dx$ = $\dfrac{1}{2}\log \left( {\sin x + \cos x} \right) + \dfrac{1}{2}x + c$
So, the correct answer is “ $\dfrac{1}{2}\log \left( {\sin x + \cos x} \right) + \dfrac{1}{2}x + c$ ”.

Note : We have many formulas related to tangent like-
$\tan 2\theta = \dfrac{{2\tan \theta }}{{1 - {{\tan }^2}\theta }}$
$\tan \left( {\alpha - \beta } \right) = \dfrac{{\tan \alpha - \tan \beta }}{{1 + \tan \alpha \tan \beta }}$
$\tan \left( {\alpha + \beta } \right) = \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }}$
We must be use the identities depending upon the nature of the questions and required answer asked.