Let $A_n = \int \tan^n x \, dx$, $\forall n \in \mathbb{N}$. If $A_{n} + A_{n-2} = \frac{\tan^{n-1} x}{n-1} + \lambda$ (where $\lambda$ is an arbitrary constant), then the value of $m$ is equal to
Step-by-Step Solution
Key Concept: Rewrite $\tan^n x + \tan^{n+2} x$ as $\tan^n x(1 + \tan^2 x) = \tan^n x \sec^2 x$ to facilitate substitution.
We need to find $A_{10} + A_{12}$ where $A_n = \int \tan^n x \, dx + \int \tan^{n+2} x \, dx = \int (\tan^n x + \tan^{n+2} x) dx = \int \tan^n x(1 + \tan^2 x) dx = \int \tan^n x \sec^2 x \, dx$. Using substitution $t = \tan x$, $dt = \sec^2 x \, dx$, we get $A_n = \int t^n dt = \frac{t^{n+1}}{n+1} = \frac{\tan^{n+1} x}{n+1}$. Therefore, $A_{10} + A_{12} = \frac{\tan^{11} x}{11} + \frac{\tan^{13} x}{13}$.
Correct Answer: B