Let $$I_{n} = \int \tan^{n}x \, dx$$ ($$n > 1$$). If $$I_{4} + I_{6} = a\tan^{5}x + bx^{5} + c$$, then the ordered pair $$(a, b)$$, is equal to

We have the family of integrals $$I_n=\int \tan^{\,n}x\,dx,\; n>1.$$

First we evaluate $$I_4=\int \tan^{4}x\,dx.$$ Because $$\tan^{2}x=\sec^{2}x-1,$$ we write

$$\tan^{4}x=(\tan^{2}x)^2=(\sec^{2}x-1)^2=\sec^{4}x-2\sec^{2}x+1.$$

So

$$I_4=\int\bigl(\sec^{4}x-2\sec^{2}x+1\bigr)\,dx =\int\sec^{4}x\,dx-2\int\sec^{2}x\,dx+\int dx.$$

To integrate $$\int\sec^{4}x\,dx$$ we set $$u=\tan x\;\;(\,du=\sec^{2}x\,dx\,).$$ Then

$$\int\sec^{4}x\,dx=\int\sec^{2}x\cdot\sec^{2}x\,dx =\int(1+u^{2})\,du=u+\frac{u^{3}}{3} =\tan x+\frac{\tan^{3}x}{3}.$$

Therefore

$$I_4=\Bigl(\tan x+\frac{\tan^{3}x}{3}\Bigr)-2(\tan x)+x+C =\frac{\tan^{3}x}{3}-\tan x+x+C.$$

Next we evaluate $$I_6=\int \tan^{6}x\,dx.$$ Using $$\tan^{6}x=(\tan^{2}x)^3=(\sec^{2}x-1)^3 =\sec^{6}x-3\sec^{4}x+3\sec^{2}x-1,$$ we obtain

$$I_6=\int\bigl(\sec^{6}x-3\sec^{4}x+3\sec^{2}x-1\bigr)\,dx =\int\sec^{6}x\,dx-3\int\sec^{4}x\,dx+3\int\sec^{2}x\,dx-\int dx.$$

For $$\int\sec^{6}x\,dx$$ we again use $$u=\tan x.$$ Because $$\sec^{6}x\,dx=\sec^{4}x\sec^{2}x\,dx=\sec^{4}x\,du,$$ and $$\sec^{4}x=(1+u^{2})^{2}=1+2u^{2}+u^{4},$$ we get

$$\int\sec^{6}x\,dx=\int\bigl(1+2u^{2}+u^{4}\bigr)\,du =u+\frac{2u^{3}}{3}+\frac{u^{5}}{5} =\tan x+\frac{2}{3}\tan^{3}x+\frac{1}{5}\tan^{5}x.$$

Hence

$$\begin{aligned} I_6&=\Bigl(\tan x+\frac{2}{3}\tan^{3}x+\frac{1}{5}\tan^{5}x\Bigr) -3\Bigl(\tan x+\frac{\tan^{3}x}{3}\Bigr) +3(\tan x)-x+C\\[4pt] &=\tan x+\frac{2}{3}\tan^{3}x+\frac{1}{5}\tan^{5}x -3\tan x-\tan^{3}x+3\tan x-x+C\\[4pt] &=\tan x-\frac{\tan^{3}x}{3}+\frac{\tan^{5}x}{5}-x+C. \end{aligned}$$

Adding the two results,

$$\begin{aligned} I_4+I_6&=\Bigl(\frac{\tan^{3}x}{3}-\tan x+x\Bigr) +\Bigl(\tan x-\frac{\tan^{3}x}{3}+\frac{\tan^{5}x}{5}-x\Bigr)+C\\[4pt] &=\frac{\tan^{5}x}{5}+C. \end{aligned}$$

Thus

$$I_{4}+I_{6}=a\tan^{5}x+bx^{5}+c =\frac{1}{5}\tan^{5}x+0\cdot x^{5}+c.$$

So $$a=\frac{1}{5}\;\;\text{and}\;\;b=0,$$ giving the ordered pair $$\left(\frac{1}{5},0\right).$$

Hence, the correct answer is Option B.