Question

How do you evaluate the integral $\int{x{{\tan }^{2}}x}$?

Answer 1

Hint: In order to solve this question we need to apply here the method of integration by parts. This means to apply \[\int{u}dv=uv-\int{v}du\]. For those we will put $u=x$ and $\dfrac{dv}{dx}={{\tan }^{2}}\left( x \right)$. After this we will use formulas ${{\tan }^{2}}\left( x \right)={{\sec }^{2}}\left( x \right)-1,\int{{{\sec }^{2}}\left( x \right)}dx=\tan x+{{c}_{1}}$ and \[\int{1}dx=x+{{c}_{2}}\] to get the right answer.

Complete step by step solution:
Consider the integral function $\int{x{{\tan }^{2}}x}$.
We will put $u=x$ and $\dfrac{dv}{dx}={{\tan }^{2}}\left( x \right)$. Now we will differentiate $u=x$ with respect to x. Therefore, we get
$\begin{align}
  & u=x \\
 & \Rightarrow \dfrac{du}{dx}=\dfrac{dx}{dx} \\
 & \Rightarrow \dfrac{du}{dx}=1 \\
\end{align}$
Now we will consider $\dfrac{dv}{dx}={{\tan }^{2}}\left( x \right)$ and write it as $dv={{\tan }^{2}}\left( x \right)dx$. After this we will use integration here. Using differentiation on both the sides of the equation $dv={{\tan }^{2}}\left( x \right)dx$ we get,
$\begin{align}
  & dv={{\tan }^{2}}\left( x \right)dx \\
 & \Rightarrow \int{dv}=\int{{{\tan }^{2}}\left( x \right)}dx \\
\end{align}$
As we know that ${{\tan }^{2}}\left( x \right)={{\sec }^{2}}\left( x \right)-1$, therefore we can write,
$\begin{align}
  & \int{dv}=\int{{{\tan }^{2}}\left( x \right)}dx \\
 & \Rightarrow v=\int{\left( {{\sec }^{2}}\left( x \right)-1 \right)}dx \\
 & \Rightarrow v=\int{{{\sec }^{2}}\left( x \right)dx}-\int{1}dx \\
\end{align}$
By taking the help of formula $\int{{{\sec }^{2}}\left( x \right)}dx=\tan x+{{c}_{1}}$ we get $v=\tan x+c-\int{1}dx$. Also, as \[\int{1}dx=x+{{c}_{2}}\],
$\begin{align}
  & v=\tan x+{{c}_{1}}-\left( x+{{c}_{2}} \right) \\
 & \Rightarrow v=\tan x+{{c}_{1}}-x-{{c}_{2}} \\
 & \Rightarrow v=\tan x-x+c\,\,\left[ \because c={{c}_{1}}-{{c}_{2}} \right] \\
\end{align}$
Now, we will use the formula of integration by parts in which we have \[\int{u}dv=uv-\int{v}du\]. After substituting all above terms in this formula we get,
\[\begin{align}
  & \int{x}{{\tan }^{2}}\left( x \right)dx=x\left( \tan x-x \right)-\int{\left( \tan x-x \right)dx} \\
 & \Rightarrow \int{x}{{\tan }^{2}}\left( x \right)dx=x\tan x-{{x}^{2}}-\left[ \int{\left( \tan x \right)dx}-\int{xdx} \right] \\
 & \Rightarrow \int{x}{{\tan }^{2}}\left( x \right)dx=x\tan x-{{x}^{2}}-\int{\left( \tan x \right)dx}+\int{xdx} \\
 & \Rightarrow \int{x}{{\tan }^{2}}\left( x \right)dx=x\tan x-{{x}^{2}}-\left( -\ln \left| \cos x \right| \right)+\dfrac{{{x}^{2}}}{2}\,\,\left[ \because \int{\left( \tan x \right)dx}=-\ln \left| \cos x \right|,\int{x}dx=\dfrac{{{x}^{2}}}{2} \right] \\
 & \Rightarrow \int{x}{{\tan }^{2}}\left( x \right)dx=x\tan x-{{x}^{2}}+\ln \left| \cos x \right|+\dfrac{{{x}^{2}}}{2} \\
 & \Rightarrow \int{x}{{\tan }^{2}}\left( x \right)dx=x\tan x+\ln \left| \cos x \right|-\dfrac{{{x}^{2}}}{2} \\
\end{align}\]
Hence, the value of \[\int{x}{{\tan }^{2}}\left( x \right)dx=x\tan x+\ln \left| \cos x \right|-\dfrac{{{x}^{2}}}{2}\].

Note:
One should always remember that there is no direct integration formula for $\int{{{\tan }^{2}}x}dx$ in trigonometry. This is why we have used the formula ${{\tan }^{2}}\left( x \right)={{\sec }^{2}}\left( x \right)-1$ and converted tangent into secant. After this we need to apply integration by parts and solve the question further. This question needs a lot of focus as there might be a chance of any mistake like, plus, minus or formula usage. We have kept $u=x$ and $\dfrac{dv}{dx}={{\tan }^{2}}\left( x \right)$ because of the format of ILATE, in which A = algebra comes first and T = trigonometric term becomes the second part for the integration by parts. There is no other method which can be applied here. So, to solve such types of questions, one should understand the complete process of integration by parts.