Question

Evaluate the following integral: $\int{\dfrac{\left( 1+\tan x \right)}{\left( 1-\tan x \right)}dx}$

Answer 1

Hint: We know that $\tan \dfrac{\pi }{4}=1$ and $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. By, using these two identities, we can convert $\dfrac{1+\tan x}{1-\tan x}=\tan \left( \dfrac{\pi }{4}+x \right)$, and can easily get the result by using the formula $\int{\tan xdx=\log \left| \sec x \right|}+c$. We can also solve this problem by replacing $\tan x=\dfrac{\sin x}{\cos x}$ and using integration by substitution method.

Complete step-by-step solution:
Let us assume the required integral to be $I=\int{\dfrac{\left( 1+\tan x \right)}{\left( 1-\tan x \right)}dx}$.
We all know very well that the tangent of angle $\dfrac{\pi }{4}$ is unity, that is
$\tan \dfrac{\pi }{4}=1$.
So, by using the above information, we can write
 $\dfrac{1+\tan x}{1-\tan x}=\dfrac{\tan \dfrac{\pi }{4}+\tan x}{1-\tan \dfrac{\pi }{4}\tan x}...\left( i \right)$
Also, we know the identity, $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$. So, by using the converse of this identity on the right hand side of equation (i), we get
$\dfrac{1+\tan x}{1-\tan x}=\tan \left( \dfrac{\pi }{4}+x \right)$
Hence, we can write
\[I=\int{\tan \left( \dfrac{\pi }{4}+x \right)dx}\].
We know that the integration of $\tan x$ is $\log \left| \sec x \right|$, that is,
$\int{\tan xdx=\log \left| \sec x \right|}+c$.
So, by using the above formula, we can write the following
\[I=\log \left| \sec \left( \dfrac{\pi }{4}+x \right) \right|+c\].
Alternatively, we can also solve this problem as shown below.
We know that $\tan x=\dfrac{\sin x}{\cos x}$. Thus, we can write
$\dfrac{1+\tan x}{1-\tan x}=\dfrac{1+\dfrac{\sin x}{\cos x}}{1-\dfrac{\sin x}{\cos x}}$
On simplifying the right hand side of this equation, we get
$\dfrac{1+\tan x}{1-\tan x}=\dfrac{\dfrac{\cos x+\sin x}{\cos x}}{\dfrac{\cos x-\sin x}{\cos x}}$
And thus, we now have
$\dfrac{1+\tan x}{1-\tan x}=\dfrac{\cos x+\sin x}{\cos x-\sin x}$
So, we can now write
$I=\int{\dfrac{\cos x+\sin x}{\cos x-\sin x}}dx$
Let us assume $t=\cos x-\sin x$.
By differentiating t, we can write
$dt=\left( -\sin x-\cos x \right)dx$
$\Rightarrow dt=-\left( \sin x+\cos x \right)dx$
Hence, we can write
$I=\int{\dfrac{-1}{t}}dt$
We know the integration identity $\int{\dfrac{1}{x}dx=\log \left| x \right|}+c$. Thus, we have
$I=-\log \left| t \right|+k$
We can now substitute the value of t back in the above equation. Hence, we get
$I=-\log \left| \cos x-\sin x \right|+k$.
Thus, we can express the required answer as \[I=\log \left| \sec \left( \dfrac{\pi }{4}+x \right) \right|+c\] or $I=-\log \left| \cos x-\sin x \right|+k$.

Note: We can see here that \[\log \left| \sec \left( \dfrac{\pi }{4}+x \right) \right|\] and $-\log \left| \cos x-\sin x \right|$ are not equivalent to each other. But these two are related by a factor of $\sqrt{2}$. Also, we know that $\log mn=\log m+\log n$. And since we know that $\log \sqrt{2}$ is a constant, these two results are equivalent to each other. So, we must never forget the constant of integration in our results.