Final answer:
The reduction formula \( \int \tan^k(x) \, dx = \frac{\tan^{k-1}(x)}{k-1} - \int \tan^{k-2}(x) \, dx \) is proved by rewriting \( \tan^k(x) \) as \( (\sec^2(x) - 1)\tan^{k-2}(x) \), setting \( u \) and \( dv \) for integration by parts, and simplifying the resulting integrals.
Step-by-step explanation:
To prove the reduction formula for \( \int \tan^k(x) \, dx = \frac{\tan^{k-1}(x)}{k-1} - \int \tan^{k-2}(x) \, dx \), we start by rewriting \( \tan^k(x) \) using the identity given: \( \tan^k(x) = (\sec^2(x) - 1)\tan^{k-2}(x) \). Then, let's set \( u = \tan^{k-1}(x) \), which implies that \( du = (k-1)\sec^2(x)\tan^{k-2}(x) \, dx \). Now integrating by parts, with \( dv = \tan^{k-2}(x)\sec^2(x) \, dx \) and \( v = \frac{\tan^{k-1}(x)}{k-1} \), we arrive at:
\( \int \tan^k(x) \, dx = \frac{\tan^{k-1}(x)}{k-1} - \int \tan^{k-2}(x)\sec^2(x) \, dx + \int \tan^{k-2}(x) \, dx \)
Since \( \sec^2(x) = 1 + \tan^2(x) \), the second integral on the right can be simplified to \( \int \tan^{k-2}(x) \, dx \), thus proving the reduction formula.