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Question: Solve \(\int {\dfrac{{1 + \tan x}}{{1 - \tan x}}} dx\)...

Solve ∫1+tan⁡x1−tan⁡xdx\int {\dfrac{{1 + \tan x}}{{1 - \tan x}}} dx

Explanation

Solution

We are asked to integrate a given function, there are many ways to integrate a given function. Here, first we use the trigonometric ratio tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{{\sin x}}{{\cos x}} in both the numerator and denominator and proceed with the simplification. After simplifying, we use the substitution method of integration to proceed further by taking cos⁡x−sin⁡x=t\cos x - \sin x = t and further simplification gives the required value.

Complete step-by-step answer:
We are asked to find the value of ∫1+tan⁡x1−tan⁡xdx\int {\dfrac{{1 + \tan x}}{{1 - \tan x}}} dx
Now let I=∫1+tan⁡x1−tan⁡xdxI = \int {\dfrac{{1 + \tan x}}{{1 - \tan x}}} dx
The function inside the integral symbol is known as the integrand.
And the derivative next to the function defines with respect to which variable we are supposed to integrate.
That is, dxdx defines that the given function should be integrated with respect to xx
Firstly, let's write down the given function
I=∫1+tan⁡x1−tan⁡xdxI = \int {\dfrac{{1 + \tan x}}{{1 - \tan x}}} dx……… (1)
We know that by trigonometric ratios that tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{{\sin x}}{{\cos x}}
Substitute tan⁡x=sin⁡xcos⁡x\tan x = \dfrac{{\sin x}}{{\cos x}} in (1) in both the numerator and denominator

Now we need to take LCM in both the numerator and denominator and proceed to simplify
I=∫cos⁡x+sin⁡xcos⁡xcos⁡x−sin⁡xcos⁡xdxI = \int {\dfrac{{\dfrac{{\cos x + \sin x}}{{\cos x}}}}{{\dfrac{{\cos x - \sin x}}{{\cos x}}}}} dx
Now we can see that cos⁡x\cos x is the common denominator for both the numerator and denominator. Hence, we can cancel it.
I=∫cos⁡x+sin⁡xcos⁡x−sin⁡xdxI = \int {\dfrac{{\cos x + \sin x}}{{\cos x - \sin x}}} dx ……(2)
Now we can use the substitution method of integration to proceed further
Take cos⁡x−sin⁡x=t\cos x - \sin x = t
We know that differentiation of cos⁡x\cos x is −sin⁡x - \sin x and differentiation of is cos⁡x\cos x
Differentiating this we get,
(−sin⁡x−cos⁡x)dx=dt\left( { - \sin x - \cos x} \right)dx = dt
Taking minus as common we get
−(sin⁡x+cos⁡x)dx=dt- \left( {\sin x + \cos x} \right)dx = dt
Multiply by −1 - 1 on both sides we get
(sin⁡x+cos⁡x)dx=−dt\left( {\sin x + \cos x} \right)dx = - dt
Now substitute cos⁡x−sin⁡x=t\cos x - \sin x = t and (sin⁡x+cos⁡x)dx=−dt\left( {\sin x + \cos x} \right)dx = - dt in (2)
∫cos⁡x+sin⁡xcos⁡x−sin⁡xdx=∫−dtt\int {\dfrac{{\cos x + \sin x}}{{\cos x - \sin x}}} dx = \int {\dfrac{{ - dt}}{t}}
Taking the minus outside the integral we get
I=−∫dttI = - \int {\dfrac{{dt}}{t}}
Now by using integration formulae, we know that ∫dxx=log⁡x+c\int {\dfrac{{dx}}{x} = \log x + c}
Using this our integral becomes
I=−∫dtt=−log⁡t+cI = - \int {\dfrac{{dt}}{t}} = - \log t + c
In rules of logarithm, we know that alog⁡b=log⁡baa\log b = \log {b^a}
Using this we get,
I=log⁡t−1+cI = \log {t^{ - 1}} + c
Now substitute cos⁡x−sin⁡x=t\cos x - \sin x = t
I=log⁡(cos⁡x−sin⁡x)−1+c I=log⁡(1cos⁡x−sin⁡x)+c  I = \log {\left( {\cos x - \sin x} \right)^{ - 1}} + c \\\ I = \log \left( {\dfrac{1}{{\cos x - \sin x}}} \right) + c \\\
We know that I=∫1+tan⁡x1−tan⁡xdxI = \int {\dfrac{{1 + \tan x}}{{1 - \tan x}}} dx
Therefore, ∫1+tan⁡x1−tan⁡xdx=log⁡(1cos⁡x−sin⁡x)+c\int {\dfrac{{1 + \tan x}}{{1 - \tan x}}} dx = \log \left( {\dfrac{1}{{\cos x - \sin x}}} \right) + c

Note: The major mistake many students make is multiplying and dividing the given fraction by the conjugate of its denominator, but it will make the problem more hectic. Same way, many get confused with the trigonometric ratios. The solution for the above problem can also be written as −log⁡(cos⁡x−sin⁡x)+c - \log \left( {\cos x - \sin x} \right) + c .This is also the correct answer.