Question

Find the integral of the function ${{\tan }^{4}}x$.

Answer 1

Hint: To find the integral of the given expression, we will first simplify the given expression into smaller expressions and then proceed ahead in our problem. The simplification can be done by breaking ${{\tan }^{4}}x$ into two factors of ${{\tan }^{2}}x$ and then using some of the trigonometric identities.

Complete step by step solution:
Let us denote the expression given to us with ‘y’, such that we need to find the integral of ‘y’ over a very small change in ‘x’. Mathematically, this could be written as:
$\begin{align}
  & \Rightarrow I=\int{y.dx} \\
 & or,I=\int{{{\tan }^{4}}x.dx} \\
\end{align}$
Where, “I” is the result of this integral.
On simplifying ‘y’, we get:
$\begin{align}
  & \Rightarrow y={{\tan }^{4}}x \\
 & \Rightarrow y={{\tan }^{2}}x\times {{\tan }^{2}}x \\
 & \Rightarrow y={{\tan }^{2}}x\times \left( {{\sec }^{2}}x-1 \right) \\
 & \Rightarrow y={{\tan }^{2}}x{{\sec }^{2}}x-{{\tan }^{2}}x \\
 & \Rightarrow y={{\tan }^{2}}x{{\sec }^{2}}x-\left( {{\sec }^{2}}x-1 \right) \\
 & \therefore y={{\tan }^{2}}x{{\sec }^{2}}x-{{\sec }^{2}}x+1 \\
\end{align}$
Now, putting the simplified value of ‘y’ in our above integral, we get:
$\begin{align}
  & \Rightarrow I=\int{\left( {{\tan }^{2}}x{{\sec }^{2}}x-{{\sec }^{2}}x+1 \right).dx} \\
 & \Rightarrow I=\int{{{\tan }^{2}}x{{\sec }^{2}}x.dx}-\int{{{\sec }^{2}}x.dx}+\int{1.dx} \\
\end{align}$
Here, let the first, second and third term of the integral respectively be ${{I}_{1}},{{I}_{2}}\text{ and }{{I}_{3}}$. Then, we have:
$\Rightarrow {{I}_{1}}=\int{{{\tan }^{2}}x{{\sec }^{2}}x.dx}$
Let us now substitute the variable ‘$\tan x$’ with the term ‘u’. Thus, the differential of ‘x’ shall also change. This can be calculated as follows:
$\begin{align}
  & \Rightarrow d\left( \tan x \right)=du \\
 & \Rightarrow {{\sec }^{2}}xdx=du \\
\end{align}$
Thus, putting the new substituted values, our new equation becomes:
$\begin{align}
  & \Rightarrow {{I}_{1}}=\int{{{u}^{2}}.du} \\
 & \Rightarrow {{I}_{1}}=\dfrac{{{u}^{3}}}{3}+{{c}_{1}} \\
\end{align}$
Re-substituting in terms of ‘x’, we get:
$\Rightarrow {{I}_{1}}=\dfrac{{{\tan }^{3}}x}{3}+{{c}_{1}}$
Secondly, we have:
$\Rightarrow {{I}_{2}}=\int{{{\sec }^{2}}x.dx}$
Since, we know the derivative of ‘$\tan x$’ is ‘${{\sec }^{2}}x$’, therefore the anti-derivative or integral of ‘${{\sec }^{2}}x$’ should be ‘$\tan x$’. Using this, we get:
$\Rightarrow {{I}_{2}}=\tan x+{{c}_{2}}$
And the last term can be calculated as:
$\begin{align}
  & \Rightarrow {{I}_{3}}=\int{1.dx} \\
 & \Rightarrow {{I}_{3}}=x+{{c}_{3}} \\
\end{align}$
Thus, putting the values of all the three integrals in the equation of “I”, we get our final result as:
$\Rightarrow I=\dfrac{{{\tan }^{3}}x}{3}+\tan x+x+C$
Where,
$\Rightarrow C={{c}_{1}}+{{c}_{2}}+{{c}_{3}}$
Hence, the integral of the function ${{\tan }^{4}}x$ comes out to be $\dfrac{{{\tan }^{3}}x}{3}+\tan x+x+C$.

Note:
Whenever using the substitution method to find out the integral of an expression, we should always be careful as to what to substitute in the place of our original variable. The substitute should be such that it reduces the complexity in our equation. This makes our new equation easy to understand and solve. The new equation sometimes might even become a standard integral just like in this case.