Question

How do you find the integral $\dfrac{{\tan (x)}}{x}$ ?

Answer 1

Hint : Here in this question, we have to find the integral for the function, the function is of trigonometry function. First, we simplify the trigonometric function and then we apply the integration to function and we obtain the required solution for the question.

Complete step-by-step answer :
In this question we have to find the integral of the function, the function is trigonometry. The cosine is one of the trigonometry ratios.
Now consider the given question $\dfrac{{\tan (x)}}{x}$ , here we can’t apply the integration directly to the given function.. So we use maclurin’s series expansion for the trigonometry function and hence we integrate the function.
Therefore the trigonometry function can be written in the form of maclurin’s series as
$\Rightarrow \tan (x) = x + \dfrac{1}{3}{x^3} + \dfrac{2}{{15}}{x^5} + ...$
Now the given function can be written as
$\Rightarrow \dfrac{{\tan (x)}}{x} = \dfrac{{x + \dfrac{1}{3}{x^3} + \dfrac{2}{{15}}{x^5} + ...}}{x}$
In the numerator take the term “x” as common and therefore the given function can be written as
$\Rightarrow \dfrac{{\tan (x)}}{x} = \dfrac{{x\left( {1 + \dfrac{1}{3}{x^2} + \dfrac{2}{{15}}{x^4} + ...} \right)}}{x}$
In the numerator and the denominator the x term will get cancels and we have
$\Rightarrow \dfrac{{\tan (x)}}{x} = \left( {1 + \dfrac{1}{3}{x^2} + \dfrac{2}{{15}}{x^4} + ...} \right)$
Now we will integrate the above function so we have
$\Rightarrow \int {\dfrac{{\tan (x)}}{x}} \,dx = \int {\left( {1 + \dfrac{1}{3}{x^2} + \dfrac{2}{{15}}{x^4} + ...} \right)} \,dx$
Integrating the each term and then we have
$\Rightarrow \int {\dfrac{{\tan (x)}}{x}} \,dx = \int {dx + \dfrac{1}{3}\int {{x^2}dx + \dfrac{2}{{15}}\int {{x^4}dx} } }$
Integrating the above function by using the integration formulas we have
$\Rightarrow \int {\dfrac{{\tan (x)}}{x}} \,dx = x + \dfrac{1}{3}.\dfrac{{{x^3}}}{3} + \dfrac{2}{{15}}.\dfrac{{{x^5}}}{5} + ... + c$
On simplifying we have
$\Rightarrow \int {\dfrac{{\tan (x)}}{x}} \,dx = x + \dfrac{{{x^3}}}{9} + \dfrac{{2{x^5}}}{{75}} + ... + c$
Hence we have integrated the given function and the c is the integration constant.
So, the correct answer is “ $x + \dfrac{{{x^3}}}{9} + \dfrac{{2{x^5}}}{{75}} + ... + c$ ”.

Note : The integration is the inverse of the differentiation. Sometimes we can’t apply the integration directly to the function because sometimes we can’t determine the solution for the given function. Here in this function we use the maclurin’s series for the tangent trigonometric function and it is given as $\tan (x) = x + \dfrac{1}{3}{x^3} + \dfrac{2}{{15}}{x^5} + ...$