Question Detail

This is a standard reduction formula for powers of tangent. Start by rewriting tanⁿ x as tanⁿ⁻² x · tan² x. Then use the trigonometric identity tan² x = sec² x - 1: I_n = ∫ tanⁿ⁻² x (sec² x - 1) dx = ∫ tanⁿ⁻² x sec² x dx - ∫ tanⁿ⁻² x dx. For the first integral, let u = tan x, so du = sec² x dx. This integral becomes ∫ uⁿ⁻² du = uⁿ⁻¹/(n-1) = (tanⁿ⁻¹ x)/(n-1). The second integral is I_(n-2). Combining these, we get I_n = (tanⁿ⁻¹ x)/(n-1) - I_(n-2).