Question

Find the integral of $\int{\tan x{{\sec }^{4}}xdx}$.

Answer 1

Hint: We need to use the theorem of substitution. We change the differential from x to another variable z using $z=f\left( x \right)$. We change all the variables in similar ways to form a new form of the integral. We solve that using basic rules of integration. At the end we go back to the initial variable form to get the final answer.

Complete step-by-step answer:
We need to find the integral value of $\int{\tan x{{\sec }^{4}}xdx}$. We convert every trigonometrical form into $\tan x$ form.
We convert ${{\sec }^{4}}x$ into $\tan x$ as ${{\sec }^{4}}x={{\left( {{\sec }^{2}}x \right)}^{2}}={{\sec }^{2}}x\times {{\sec }^{2}}x={{\sec }^{2}}x\left( 1+{{\tan }^{2}}x \right)$.
Now using conversion of differential from $dx$ to $dz$ where $z=\tan x$.
We take the differential form on the both side and get
\[\begin{align}
  & z=\tan x \\
 & \Rightarrow dz={{\sec }^{2}}xdx \\
\end{align}\]
We got two equation $z=\tan x$, \[dz={{\sec }^{2}}xdx\] and ${{\sec }^{4}}x={{\sec }^{2}}x\left( 1+{{\tan }^{2}}x \right)$. We place these in the main equation to reform the integral as
$\begin{align}
  & \int{\tan x{{\sec }^{4}}xdx} \\
 & =\int{\left( z \right)\left\{ {{\sec }^{2}}x\left( 1+{{\tan }^{2}}x \right) \right\}dx} \\
 & =\int{z\left( 1+{{\left( z \right)}^{2}} \right)\left( {{\sec }^{2}}xdx \right)} \\
 & =\int{z\left( 1+{{z}^{2}} \right)dz} \\
 & =\int{\left( z+{{z}^{3}} \right)dz} \\
\end{align}$
We converted the integral from x to z.
Now we apply the rules of integral to find the solution of the problem.
We know that \[\int{{{x}^{n}}}dx=\dfrac{{{x}^{n+1}}}{n+1}+c\].
So, $\int{\left( z+{{z}^{3}} \right)dz}=\int{zdz}+\int{{{z}^{3}}dz}=\dfrac{{{z}^{2}}}{2}+\dfrac{{{z}^{4}}}{4}+c$.
As the given integral was in the form of x, we need to go back to x from z.
We have the relation of $z=\tan x$.
So, the integral becomes $\dfrac{{{\left( \tan x \right)}^{2}}}{2}+\dfrac{{{\left( \tan x \right)}^{4}}}{4}+c$. Here c is the integral constant.
So, the integral of $\int{\tan x{{\sec }^{4}}xdx}$ is $\dfrac{{{\left( \tan x \right)}^{2}}}{2}+\dfrac{{{\left( \tan x \right)}^{4}}}{4}+c$.

Note: We need to remember that the method of substitution is not absolutely necessary for solving the problem as we can solve it using the trigonometric form also. But in that case the form becomes very tough to find in its differential form as we have to deal with the $d\left( \tan x \right)$. So, to make the problem easy and understandable we use a method of substitution.