Question

How do you find the integral of $\int{{{\tan }^{n}}xdx}$ if n is an integer?

Answer 1

Hint: We try to form the integral in its recurring form of power. The power reduces by 2 in every integral. We use the identity for ${{\tan }^{2}}x={{\sec }^{2}}x-1$. For even and odd value of n the end of the integral differs.

Complete step by step answer:
It’s given that n is an integer. This means $n\in \mathbb{Z}$.
We need to find the value of $\int{{{\tan }^{n}}xdx}$. Let’s assume ${{I}_{n}}=\int{{{\tan }^{n}}xdx}$.
We take the trigonometric identity ${{\tan }^{2}}x={{\sec }^{2}}x-1$.
We break the power of n in $\left( n-2 \right)$ and 2.
We get ${{\tan }^{n}}x={{\tan }^{n-2}}x.{{\tan }^{2}}x$. We replace the value of ${{\tan }^{2}}x$.
Therefore, ${{\tan }^{n}}x={{\tan }^{n-2}}x\left( {{\sec }^{2}}x-1 \right)={{\tan }^{n-2}}x{{\sec }^{2}}x-{{\tan }^{n-2}}x$.
Therefore, \[{{I}_{n}}=\int{{{\tan }^{n}}xdx}=\int{{{\tan }^{n-2}}x{{\sec }^{2}}xdx}-\int{{{\tan }^{n-2}}xdx}\].
Now as ${{I}_{n}}=\int{{{\tan }^{n}}xdx}$, we can say $\int{{{\tan }^{n-2}}xdx}={{I}_{n-2}}$.
Now we find the integral value of \[\int{{{\tan }^{n-2}}x{{\sec }^{2}}xdx}\]. We take $\tan x=z$.
Differentiating both sides of the equation of $\tan x=z$, we get
$\begin{align}
  & \dfrac{d}{dx}\left( \tan x \right)=\dfrac{d}{dx}\left( z \right) \\
 & \Rightarrow {{\sec }^{2}}x=\dfrac{dz}{dx} \\
 & \Rightarrow {{\sec }^{2}}xdx=dz \\
\end{align}$
Now we replace the values of the integral \[\int{{{\tan }^{n-2}}x{{\sec }^{2}}xdx}\] and get
\[\int{{{\tan }^{n-2}}x{{\sec }^{2}}xdx}=\int{{{\left( \tan x \right)}^{n-2}}\left( {{\sec }^{2}}xdx \right)}=\int{{{z}^{n-2}}dz}\].
We know that the integral of ${{n}^{th}}$ power of x is \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c\].
Therefore, \[\int{{{z}^{n-2}}dz}=\dfrac{{{z}^{n-2+1}}}{n-2+1}+c=\dfrac{{{z}^{n-1}}}{n-1}+c\].
We replace the value of z and get \[\int{{{\tan }^{n-2}}x{{\sec }^{2}}xdx}=\dfrac{{{\tan }^{n-1}}x}{n-1}+c\].
So, the total integral \[{{I}_{n}}=\int{{{\tan }^{n}}xdx}=\int{{{\tan }^{n-2}}x{{\sec }^{2}}xdx}-\int{{{\tan }^{n-2}}xdx}\] becomes
\[{{I}_{n}}=\int{{{\tan }^{n}}xdx}=\dfrac{{{\tan }^{n-1}}x}{n-1}-{{I}_{n-2}}+c\]
Now we can see the integral power of ratio tan reducing in every integral by 2.
Therefore, the final integral solution is dependent on the value of n.
The next step of the integral is the \[{{I}_{n-2}}\]. Replacing value of $\left( n-2 \right)$ in the value of n, we get
\[{{I}_{n-2}}=\int{{{\tan }^{n-2}}xdx}=\dfrac{{{\tan }^{n-3}}x}{n-3}-{{I}_{n-4}}+c\].
Now replacing value of \[{{I}_{n-2}}\] in \[{{I}_{n}}\], we get \[{{I}_{n}}=\dfrac{{{\tan }^{n-1}}x}{n-1}-\left( \dfrac{{{\tan }^{n-3}}x}{n-3}-{{I}_{n-4}} \right)+c=\dfrac{{{\tan }^{n-1}}x}{n-1}-\dfrac{{{\tan }^{n-3}}x}{n-3}+{{I}_{n-4}}+c\].
We can see the sign is changing in rotation as plus then minus and then plus.
So, for value of n being even the integral value is
\[{{I}_{n}}=\dfrac{{{\tan }^{n-1}}x}{n-1}-\dfrac{{{\tan }^{n-3}}x}{n-3}+....+\tan x+{{\left( -1 \right)}^{\dfrac{n}{2}}}x+c\].
For value of n being odd the integral value is
\[{{I}_{n}}=\dfrac{{{\tan }^{n-1}}x}{n-1}-\dfrac{{{\tan }^{n-3}}x}{n-3}+....+\dfrac{{{\tan }^{2}}x}{2}+{{\left( -1 \right)}^{\dfrac{n-1}{2}}}\log \left| \sec x \right|+c\].

Note: The constant in the integral has been kept at c at all the time. We are taking the final constant as c. Changing the constant all the time would have been difficult to operate. Also we need to remember instead of using the final formula we should always use the step by step integral to avoid the confusion of signs.