Let $$I\left(m,n\right)=\int_0^m\tan^nx\ dx$$. What is the value of $$2I\left(\frac{\ \pi\ }{4},2\right)+3I\left(\frac{\ \pi\ }{4},3\right)+2I\left(\frac{\ \pi\ }{4},4\right)+3I\left(\frac{\ \pi\ }{4},5\right)$$?

To evaluate the expression efficiently, let us use the standard reduction formula for $$I(n) = \int_0^{\pi/4} \tan^n x \, dx$$.

We know that for $$n \ge 2$$:

$$I(n) + I(n-2) = \int_0^{\pi/4} (\tan^n x + \tan^{n-2} x) \, dx = \int_0^{\pi/4} \tan^{n-2} x (\sec^2 x) \, dx$$

Using substitution $$u = \tan x$$, $$du = \sec^2 x \, dx$$:

$$I(n) + I(n-2) = \left[ \frac{\tan^{n-1} x}{n-1} \right]_0^{\pi/4} = \frac{1}{n-1}$$

Thus, we have the identity:

$$I(n) + I(n-2) = \frac{1}{n-1}$$

We need to find the value of:

$$S = 2I\left(\frac{\pi}{4}, 2\right) + 3I\left(\frac{\pi}{4}, 3\right) + 2I\left(\frac{\pi}{4}, 4\right) + 3I\left(\frac{\pi}{4}, 5\right)$$

Let us group the terms using $$I(n) + I(n-2) = \frac{1}{n-1}$$:

• Group 1: $$2I(2) + 2I(4) = 2 \left( I(4) + I(2) \right) = 2 \left( \frac{1}{4-1} \right) = 2 \left( \frac{1}{3} \right) = \frac{2}{3}$$
  
• Group 2: $$3I(3) + 3I(5) = 3 \left( I(5) + I(3) \right) = 3 \left( \frac{1}{5-1} \right) = 3 \left( \frac{1}{4} \right) = \frac{3}{4}$$
  

Adding these two results together:

$$S = \frac{2}{3} + \frac{3}{4} = \frac{8 + 9}{12} = \frac{17}{12}$$

The correct option is B.