Question

If \[\int {{{\tan }^4}xdx = a{{\tan }^3}x + b\tan x + \phi (x)} \] then which of the following is true,
A) \[a = \dfrac{1}{2}\]
B) \[b = 1\]
C) \[\phi (x) = x + C\]
D) \[b = - 1\]

Answer 1

Hint: To solve this problem, we need some integration formulae, they are listed below,
\[\int {{{\sec }^2}xdx = \tan x + C} \]
\[\int {dx = x + C} \]

Complete step-by-step answer:
It is given in the question that, \[\int {{{\tan }^4}xdx = a{{\tan }^3}x + b\tan x + \phi (x)} \]
At first, we integrate the left hand side of the equation which is nothing but\[\int {{{\tan }^4}xdx} \]
Let us consider\[\int {{{\tan }^4}xdx} \]
On solving the above term we get,
\[\int {{{\tan }^4}xdx} \]\[ = \int {{{\tan }^2}x.{{\tan }^2}xdx} \]
Now let us rewrite the above expression as,
\[\int {{{\tan }^2}x.{{\tan }^2}xdx} \]\[ = \int {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}.{{\tan }^2}xdx} \]
Let us solve the above equation using a trigonometric identity then we get,
\[\int {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}.{{\tan }^2}xdx} \]\[ = \int {\dfrac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}}.{{\tan }^2}xdx} \]
Again the above expression can be written as,
\[\int {\dfrac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}}.{{\tan }^2}xdx} \]\[ = \int {\dfrac{{{{\tan }^2}x}}{{{{\cos }^2}x}}dx} - \int {{{\tan }^2}xdx} \]…. (1)
We know the following differentiation formula, \[d[\tan x] = {\sec ^2}xdx\]
As we know the relation between cosines and secant function the above equation is written as
\[d[\tan x] = \dfrac{{dx}}{{{{\cos }^2}x}}\]
Substituting the above formula in equation (1) we get,
\[\int {{{\tan }^4}xdx} \]\[ = \int {{{\tan }^2}xd[\tan x} ] - \int {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}dx} \]
By integrating the first term in the right hand side of the above equation we get,
\[\int {{{\tan }^4}xdx} \]\[ = \dfrac{{{{\tan }^3}x}}{3} - \int {\dfrac{{1 - {{\cos }^2}x}}{{{{\cos }^2}x}}dx} \]
Now let us solve the integration in the second term of the equation in the right hand side we get,
\[\int {{{\tan }^4}xdx} \]\[ = \dfrac{{{{\tan }^3}x}}{3} - \int {{{\sec }^2}xdx + \int {dx} } \]
From the formula given in the hint we can integrate the terms in the above equation, therefore we get,
\[\int {{{\tan }^4}xdx} \]\[ = \dfrac{{{{\tan }^3}x}}{3} - \tan x + x + C\]
Now let us compare the above equation with the question we get,
\[\dfrac{{{{\tan }^3}x}}{3} - \tan x + x + C\] with \[a{\tan ^3}x + b\tan x + \phi (x)\]
Hence we have compared the equation as a result we can find the value of a, b and c.
The values of a, b and c are\[a = \dfrac{1}{3}\], \[b = - 1\] and \[\phi (x) = x + C\]respectively.

Hence, From the given options we have found the correct options are (C) and (D) \[\phi (x) = x + C\], \[b = - 1\] respectively.

Note:
Here, two options are correct. And we have used the following trigonometric identities
${\sin ^2}\theta = 1 - {\cos ^2}\theta {\text{ \& }}{\sec ^2}\theta = \dfrac{1}{{{\text{co}}{{\text{s}}^2}\theta }}$