Evaluate the following integral ∫〖〖tan〗^3 x〗 〖sec〗^5 x dx

∫〖〖tan〗^3 x〗  〖sec〗^5 x dx=∫〖〖tan〗^2 x〗  〖sec〗^4 x(sec⁡x  tan⁡x )  dx
=∫〖(〖sec〗^2 x-1) 〖sec〗^4 x〗 (sec⁡x  tan⁡x )  dx
Substituting  u=sec⁡x   ,then du=sec⁡x  tan⁡x  dx,we obtain:
∫〖〖tan〗^3 x〗  〖sec〗^5 x dx=∫〖(u^2-1) u^4 〗 du
=∫(u^6-u^4 )  du
=1/7 u^7-1/5 u^5+c
=1/7 〖sec〗^7 x-1/5 〖sec〗^5 x+c