If the integral is given as ${{I}_{m}}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m}}dx}$, then ${{I}_{3}}+{{I}_{5}}+{{I}_{7}}+{{I}_{9}}$ is equal to
(A) $\dfrac{3}{8}$
(B) $\dfrac{3}{7}$
(C) $\dfrac{2}{5}$
(D) $\dfrac{4}{9}$

Hint: We solve this question by first writing ${{\left( \tan x \right)}^{m}}$ as ${{\left( \tan x \right)}^{m-2}}\times {{\left( \tan x \right)}^{2}}$ in the integral ${{I}_{m}}$. Then we use the formula ${{\sec }^{2}}x-{{\tan }^{2}}x=1$ and write ${{I}_{m}}$ in terms of ${{I}_{m-2}}$. Then we integrate the remaining term by assuming $\tan x=t$ and then converting the integral in terms of t and find the value of integral using the formula $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$. Then we get a relation between ${{I}_{m}}$ and ${{I}_{m-2}}$. Then we substitute the values m=5 and m=9 and then adding them we get the required value.

Complete step-by-step solution:
Let us consider the given integral, ${{I}_{m}}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m}}dx}$
Let us consider the formula,
${{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}$
Then we can write ${{\left( \tan x \right)}^{m}}$ as
$\begin{align}
& {{\left( \tan x \right)}^{m}}={{\left( \tan x \right)}^{m-2}}\times {{\left( \tan x \right)}^{2}} \\
 & {{\left( \tan x \right)}^{m}}={{\left( \tan x \right)}^{m-2}}\times {{\tan }^{2}}x \\
\end{align}$
So, we can write the integral as,
${{I}_{m}}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( {{\left( \tan x \right)}^{m-2}}\times {{\tan }^{2}}x \right)dx}$
Now, let us consider the formula for trigonometric identity,
$\begin{align}
  & \Rightarrow {{\sec }^{2}}x-{{\tan }^{2}}x=1 \\
 & \Rightarrow {{\tan }^{2}}x={{\sec }^{2}}x-1 \\
\end{align}$
Using that we can write the integral as,
 $\begin{align}
  & \Rightarrow {{I}_{m}}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( {{\left( \tan x \right)}^{m-2}}\times \left( {{\sec }^{2}}x-1 \right) \right)dx} \\
 & \Rightarrow {{I}_{m}}=\int\limits_{0}^{\dfrac{\pi }{4}}{\left( {{\left( \tan x \right)}^{m-2}}{{\sec }^{2}}x-{{\left( \tan x \right)}^{m-2}} \right)dx} \\
 & \Rightarrow {{I}_{m}}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m-2}}{{\sec }^{2}}xdx}-\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m-2}}dx}.............\left( 1 \right) \\
\end{align}$
But we have that ${{I}_{m}}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m}}dx}$, so we can write $\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m-2}}dx}$ as
${{I}_{m-2}}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m-2}}dx}$
Substituting this value in the equation (1), we get
\[\begin{align}
  & \Rightarrow {{I}_{m}}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m-2}}{{\sec }^{2}}xdx}-{{I}_{m-2}} \\
 & \Rightarrow {{I}_{m}}+{{I}_{m-2}}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m-2}}{{\sec }^{2}}xdx}..........\left( 2 \right) \\
\end{align}\]
Now let us consider the integral on the right-hand side of the equation (2),
\[\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m-2}}{{\sec }^{2}}xdx}\]
Let $\tan x=t$.
Let us consider the formula for differentiating \[d\left( \tan x \right)={{\sec }^{2}}xdx\]
Then differentiating it using the above formula we get, ${{\sec }^{2}}xdx=dt$
When $x=0$, we get $t=\tan 0=0$.
When $x=\dfrac{\pi }{4}$, we get $t=\tan \dfrac{\pi }{4}=1$.
So, changing the variables in the integral from x to t we get,
\[\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m-2}}{{\sec }^{2}}xdx}=\int\limits_{0}^{1}{{{t}^{m-2}}dt}\]
Now let us consider the formula for integration,
$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}$
Using the above formula, we can solve our integral as,
\[\begin{align}
  & \Rightarrow \int\limits_{0}^{1}{{{t}^{m-2}}dt}=\left. \dfrac{{{t}^{m-1}}}{m-1} \right]_{0}^{1} \\
 & \Rightarrow \int\limits_{0}^{1}{{{t}^{m-2}}dt}=\left( \dfrac{{{1}^{m-1}}}{m-1}-\dfrac{{{0}^{m-1}}}{m-1} \right) \\
 & \Rightarrow \int\limits_{0}^{1}{{{t}^{m-2}}dt}=\left( \dfrac{1}{m-1}-\dfrac{0}{m-1} \right)=\dfrac{1}{m-1} \\
\end{align}\]
So, we get
\[\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m-2}}{{\sec }^{2}}xdx}=\int\limits_{0}^{1}{{{t}^{m-2}}dt}=\dfrac{1}{m-1}\]
Substituting this value in equation (2), we get
\[\Rightarrow {{I}_{m}}+{{I}_{m-2}}=\int\limits_{0}^{\dfrac{\pi }{4}}{{{\left( \tan x \right)}^{m-2}}{{\sec }^{2}}xdx}=\dfrac{1}{m-1}\]
So, we get that for any $m$,
\[\Rightarrow {{I}_{m}}+{{I}_{m-2}}=\dfrac{1}{m-1}..........\left( 3 \right)\]
Now we need to find the value of ${{I}_{3}}+{{I}_{5}}+{{I}_{7}}+{{I}_{9}}$. Now let us divide it into two parts ${{I}_{3}}+{{I}_{5}}$ and ${{I}_{7}}+{{I}_{9}}$
Now let us consider ${{I}_{3}}+{{I}_{5}}$. I it similar to the value in the left-hand side of equation (3).
So, let us substitute the value $m=5$, then we get,
\[\begin{align}
  & \Rightarrow {{I}_{5}}+{{I}_{5-2}}=\dfrac{1}{5-1} \\
 & \Rightarrow {{I}_{5}}+{{I}_{3}}=\dfrac{1}{4}..............\left( 4 \right) \\
\end{align}\]
Now let us consider ${{I}_{7}}+{{I}_{9}}$ and as it is also similar to equation (3), let us substitute the value $m=9$, then we get,
 \[\begin{align}
  & \Rightarrow {{I}_{9}}+{{I}_{9-2}}=\dfrac{1}{9-1} \\
 & \Rightarrow {{I}_{9}}+{{I}_{7}}=\dfrac{1}{8}..............\left( 5 \right) \\
\end{align}\]
So, adding the equations (4) and (5) we get,
$\begin{align}
  & \Rightarrow {{I}_{3}}+{{I}_{5}}+{{I}_{7}}+{{I}_{9}}=\dfrac{1}{4}+\dfrac{1}{8} \\
 & \Rightarrow {{I}_{3}}+{{I}_{5}}+{{I}_{7}}+{{I}_{9}}=\dfrac{3}{8} \\
\end{align}$
Hence, we get the value of ${{I}_{3}}+{{I}_{5}}+{{I}_{7}}+{{I}_{9}}$ as $\dfrac{3}{8}$.
Hence, the answer is Option A.

Note: We can also solve the value of ${{I}_{3}}$, ${{I}_{5}}$, ${{I}_{7}}$ and ${{I}_{9}}$ separately and add them to find the answer also. But it is a long and hectic way of solving the problem. The common mistake one makes is one might forget to change the limits when the variable is changed from x to t.