Question

What is the integral of \[\int {{{\tan }^3}\left( x \right)dx} \] ?

Answer 1

Hint: Here in this question given an Indefinite integral, we have to find the integrated value of the given trigonometric function. This can be solved by the substitution method and later integrated by using the standard trigonometric formula of integration. And by further simplification we get the required solution.

Complete step by step solution:
Integration is the inverse process of differentiation. An integral which is not having any upper and lower limit is known as an indefinite integral.
Consider the given function.
 \[ \Rightarrow \int {{{\tan }^3}\left( x \right)dx} \]
It can be rewritten as:
 \[ \Rightarrow \int {{{\tan }^2}\left( x \right)\tan \left( x \right)dx} \] -------(1)
Now, using the one of the standard identity of trigonometry i.e., \[{\tan ^2}\left( \theta \right) + 1 = {\sec ^2}\left( \theta \right)\]
On rearranging we can also written as: \[{\tan ^2}\left( \theta \right) = {\sec ^2}\left( \theta \right) - 1\]
On substituting \[{\tan ^2}\left( x \right)\] in equation (1), we have
 \[ \Rightarrow \int {\left( {{{\sec }^2}\left( x \right) - 1} \right)\tan \left( x \right)dx} \]
 \[ \Rightarrow \int {\left( {{{\sec }^2}\left( x \right)\tan \left( x \right) - \tan \left( x \right)} \right)dx} \]
Integrate each term with respect to x, then
 \[ \Rightarrow \int {{{\sec }^2}\left( x \right)\tan \left( x \right)dx} - \int {\tan \left( x \right)dx} \] ---------(2)
Now, consider
 \[ \Rightarrow \int {{{\sec }^2}\left( x \right)\tan \left( x \right)dx} \] --------(3)
Apply the Substitution put \[u = \tan \left( x \right)\] then
On differentiating u with respect to x, we get
 \[\dfrac{{du}}{{dx}} = {\sec ^2}\left( x \right)\]
Or it can be written as
 \[du = {\sec ^2}\left( x \right)dx\]
Then, Integral (2) becomes
 \[ \Rightarrow \int {udu} \]
As we know, integration of \[\int {xdx = \dfrac{{{x^2}}}{2} + c} \], then
 \[ \Rightarrow \dfrac{{{u^2}}}{2} + {c_1}\]
Substitute the u value, then
 \[ \Rightarrow \dfrac{{{{\tan }^2}\left( x \right)}}{2} + {c_1}\] ----------(a)
Now, consider
 \[ \Rightarrow \int {\tan \left( x \right)} dx\]
By the definition \[\tan x = \dfrac{{\sin x}}{{\cos x}}\], then
 \[ \Rightarrow \int {\dfrac{{\sin \left( x \right)}}{{\cos \left( x \right)}}} dx\] -----------(4)
By substitution method, put \[v = \cos \left( x \right)\] . Then
On differentiating v with respect to x, we get
 \[\dfrac{{dv}}{{dx}} = \sin \left( x \right)\]
Or it can be written as
 \[dv = \sin \left( x \right)dx\]
Then, Integral (3) becomes
 \[ \Rightarrow \int {-\dfrac{1}{v}dv} \]
As we know, integration of \[\int {\dfrac{1}{x}dx = \ln \left| x \right| + c} \], then
 \[ \Rightarrow \ln \left| v \right| + {c_2}\]
Substitute the v value, then
 \[ \Rightarrow \ln \left| {\cos \left( x \right)} \right| + {c_2}\] ----------(b)
Substitute equation (a) and (b) in equation (2), then
 \[ \Rightarrow \dfrac{{{{\tan }^2}\left( x \right)}}{2} + {c_1} + \ln \left| {\cos \left( x \right)} \right| + {c_2}\]
Where, \[{c_1}\] and \[{c_2}\] are integrating constant on adding both we can write \[C\] as integrating constant.
 \[ \Rightarrow \dfrac{{{{\tan }^2}\left( x \right)}}{2} + \ln \left| {\cos \left( x \right)} \right| + C\]
Hence, it’s a required solution.
So, the correct answer is “ \[ \dfrac{{{{\tan }^2}\left( x \right)}}{2} + \ln \left| {\cos \left( x \right)} \right| + C\] ”.

Note: By simplifying the question using the substitution we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So, simplification is needed. We must know the differentiation and integration formulas. The standard integration formulas for the trigonometric ratios must know.