Question

Evaluate: $\int\limits_0^{\dfrac{\pi }{4}} {\log \left( {1 + \tan x} \right)} dx$?

Answer 1

The given integral is a definite integral so, we will get a unique value for this integral. We will first let the given integral be equal to $I$. Further, we will use the properties of definite integrals to simplify the question. Then we will use the difference formula for tangent function. At last, we will use the logarithmic property.

Formula used:
$\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx$
$\Rightarrow \tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$

Complete step by step answer:
$I = \int\limits_0^{\dfrac{\pi }{4}} {\log \left( {1 + \tan x} \right)} dx$
Now using the property of definite integral that $\int\limits_0^a {f\left( x \right)} dx = \int\limits_0^a {f\left( {a - x} \right)} dx$, we get;
$\Rightarrow I = \int\limits_0^{\dfrac{\pi }{4}} {\log \left[ {1 + \tan \left( {\dfrac{\pi }{4} - x} \right)} \right]} dx$

Now we will use the difference formula form tangent function i.e., $\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$, we get,
$\Rightarrow I = \int\limits_0^{\dfrac{\pi }{4}} {\log \left[ {1 + \dfrac{{\tan \dfrac{\pi }{4} - \tan x}}{{1 + \tan \dfrac{\pi }{4}\tan x}}} \right]} dx$
Now we will put the value of $\tan \dfrac{\pi }{4}$. So, we have;
$\Rightarrow I = \int\limits_0^{\dfrac{\pi }{4}} {\log \left[ {1 + \dfrac{{1 - \tan x}}{{1 + \tan x}}} \right]} dx$
Now we will take the LCM and solve it. So, we get;
$\Rightarrow I = \int\limits_0^{\dfrac{\pi }{4}} {\log \left[ {\dfrac{{1 + \tan x + 1 - \tan x}}{{1 + \tan x}}} \right]} dx$

Cancelling the terms, we get;
$\Rightarrow I = \int\limits_0^{\dfrac{\pi }{4}} {\log \left[ {\dfrac{2}{{1 + \tan x}}} \right]} dx$
Now by using the property of logarithm that;
$\log \dfrac{a}{b} = \log a - \log b$
We get;
$\Rightarrow I = \int\limits_0^{\dfrac{\pi }{4}} {\log 2} dx - \int\limits_0^{\dfrac{\pi }{4}} {\log \left( {1 + \tan x} \right)} dx$
Now we can see that the second term on the RHS is the same expression we have assumed to be equal to $I$. So, replacing we get;
$\Rightarrow I = \int\limits_0^{\dfrac{\pi }{4}} {\log 2} dx - I$

Now shifting $I$, to the LHS we get
$\Rightarrow 2I = \int\limits_0^{\dfrac{\pi }{4}} {\log 2} dx$
Taking the logarithmic term out of integration as it is a constant, we get;
$\Rightarrow 2I = \log 2\int\limits_0^{\dfrac{\pi }{4}} {dx}$
After integration we get;
$\Rightarrow 2I = \log 2\left[ x \right]_0^{\dfrac{\pi }{4}}$
Putting the limits, we get;
$\Rightarrow 2I = \log 2\left( {\dfrac{\pi }{4} - 0} \right)$
On solving we get;
$\therefore I = \dfrac{\pi }{8}\log 2$

Therefore, the integration of $\int\limits_0^{\dfrac{\pi }{4}} {\log \left( {1 + \tan x} \right)} dx$ is $\dfrac{\pi }{8}\log 2$.

Note: One thing to note is that the value we have got for this integral is the area under the curve $\log \left( {1 + \tan x} \right)$ from $x = 0$ to $x = \dfrac{\pi }{4}$. We can also solve this question by drawing the curve and finding the area for the given limit. But as it is somewhat difficult to draw the curve for the given expression, we have simply found the value by integration. But, in some other cases where drawing a graph will be easier, we can check by finding the area.