Question

Integration = x³ tan⁴ x⁴ sec² x⁴

Answer 1

Answer:

$\frac{tan^{5} {x}^{4} }{15} + c$

Step-by-step explanation:

$\int {x}^{3} tan^{4} {x}^{4}sec^{2} {x}^{4} dx \\ let \: \: y = {x}^{4} \\ dy = 4 {x}^{3} dx \\ \frac{dy}{3} = {x}^{3} dx \\ now \: \: \: \frac{1}{3} \int \: tan^{4} y \: sec^{2}ydx \\ again \: \: let \: \: z = tany \\ dz = sec^{2} ydy \\ now \: \: \: \: \frac{1}{3} \int \: {z}^{4} dz = \frac{1}{3} \times \frac{ {z}^{5} }{5} = \frac{ {z}^{5} }{15} + c \\ \frac{tan^{5}y }{15} + c = \frac{tan^{5} {x}^{4} }{15} + c$