If $\int \tan ^4 x \mathrm{dx}=a \tan ^3 x+\mathrm{b} \tan x+\mathrm{c} x+\mathrm{k}$ (where k is the constant of integration) then the value of $a-b+c=$

Consider the integral I = $\int \tan^4 x dx$. Rewriting using the identity $\tan^2 x = \sec^2 x - 1$:

$I = \int \tan^2 x (\sec^2 x - 1) dx = \int \tan^2 x \sec^2 x dx - \int \tan^2 x dx$

For the first term, substitute $u = \tan x$, $du = \sec^2 x dx$:

$\int u^2 du = \frac{u^3}{3} = \frac{\tan^3 x}{3}$

For the second term, apply the identity again:

$\int \tan^2 x dx = \int (\sec^2 x - 1) dx = \tan x - x$

The complete integral becomes:

$I = \frac{\tan^3 x}{3} - \tan x + x + k$

Matching with the given form $a \tan^3 x + b \tan x + c x + k$ yields coefficients $a = \frac{1}{3}$, $b = -1$, $c = 1$. Therefore:

$a - b + c = \frac{1}{3} - (-1) + 1 = \frac{7}{3}$