Question

Find the value of $\int {x{{\tan }^2}xdx}$.

Answer 1

Here, we will first rewrite the integrand of the given integral function using a suitable trigonometric identity. We will then apply the integration by parts formula to simplify the integrand. Then we will use the suitable integration formula to integrate the given function.

Formula Used:
We will use the following formulas:

1. Trigonometric Identity: $1 + {\tan ^2}\theta = {\sec ^2}\theta$
2. Integration by Parts: $\int {uvdx = uv - \int {vdu} }$
3. Derivative Formula: $\dfrac{d}{{dx}}\left( C \right) = 1$
4. Integral Formula: $\int {{{\sec }^2}xdx = \tan x}$, $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$ and $\int {\dfrac{{\sin x}}{{\cos x}}dx = - \ln \left( {\cos x} \right) + C}$

Complete step by step solution:
We are given an Integral function $\int {x{{\tan }^2}xdx}$.
We know that $1 + {\tan ^2}\theta = {\sec ^2}\theta$, so we can write
${\tan ^2}\theta = {\sec ^2}\theta - 1$
Substituting this value in the given function, we get
$\Rightarrow \int {x{{\tan }^2}xdx} = \int {x\left( {{{\sec }^2}x - 1} \right)dx}$.
Now, by multiplying the terms in the integrand, we get
$\Rightarrow \int {x{{\tan }^2}xdx} = \int {\left( {x{{\sec }^2}x - x} \right)dx}$
By segregating the integrand in the integral function, we get
$\Rightarrow \int {x{{\tan }^2}xdx} = \int {x{{\sec }^2}xdx - \int {xdx} }$
Now, by using Integration by Parts, $\int {uvdx = uv - \int {vdu} }$, for the first Integral function, we get $u = x$ according to ILATE rule and $v = {\sec ^2}x$ .
Now, we will differentiate the variable $u$, so we get
$du = dx$…………………………….$\left( 1 \right)$
Now, we will integrate the variable $v$ using the formula $\int {{{\sec }^2}xdx = \tan x}$, so we get
$\int {{{\sec }^2}xdx = \tan x}$ …………………………………………$\left( 2 \right)$
Now, by substituting equation$\left( 1 \right)$ and $\left( 2 \right)$ in the integration by parts formula, we get
$\int {x{{\sec }^2}xdx = x\tan x - \int {\tan xdx} }$ …………………………………………..$\left( 3 \right)$
Now, by substituting equation $\left( 3 \right)$ in the integral function, we get
$\Rightarrow \int {x{{\tan }^2}xdx} = x\tan x - \int {\tan xdx - \int {xdx} }$
$\Rightarrow \int {x{{\tan }^2}xdx} = x\tan x - \int {\dfrac{{\sin x}}{{\cos x}}dx - \int {xdx} }$
Now, we will integrate the function by using the integral formula $\int {{x^n}dx = \dfrac{{{x^{n + 1}}}}{{n + 1}}}$ and $\int {\dfrac{{\sin x}}{{\cos x}}dx = - \ln \left( {\cos x} \right) + C}$, we get
$\Rightarrow \int {x{{\tan }^2}xdx} = x\tan x - \left[ { - \ln \left( {\cos x} \right)} \right] - \dfrac{{{x^2}}}{2} + C$
$\Rightarrow \int {x{{\tan }^2}xdx} = x\tan x - \left[ {\ln \left( {{{\cos }^{ - 1}}x} \right)} \right] - \dfrac{{{x^2}}}{2} + C$
Using the Inverse Trigonometric ratio $\dfrac{1}{{\cos x}} = \sec x$, we get
$\Rightarrow \int {x{{\tan }^2}xdx} = x\tan x - \left[ {\ln \left( {\sec x} \right)} \right] - \dfrac{{{x^2}}}{2} + C$

Therefore, the value of $\int {x{{\tan }^2}xdx}$ is $x\tan x - \left[ {\ln \left( {\sec x} \right)} \right] - \dfrac{{{x^2}}}{2} + C$.

Note:
We know that Integration is the process of adding small parts to find the whole parts. While performing the Integration by Parts, the first function is selected according to the ILATE rule where Inverse Trigonometric function, followed by Logarithmic function, Arithmetic Function, Trigonometric Function and at last Exponential Function. Integration by Parts is applicable only when the integrand is a product of two Functions.