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

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
${\sin ^2}\theta = 1 - {\cos ^2}\theta {\text{ \& }}{\sec ^2}\theta = \dfrac{1}{{{\text{co}}{{\text{s}}^2}\theta }}$