Question

Evaluate: ∫ tan3 x .sec6x dx

Answer 1

= ∫ tan3 x (1 + tan2 x)2 sec2 x dx

= ∫ t3 (1 + t2)2 dt,

Let tan x = t

= sec2 x dx = dt

= ∫ t3 (t4 + 2t2 + 1) dt

= ∫ (t7+ 2t5 + t3) dt

= ((t8/ 8) + (t3/ 3) + (t4/ 4)) + c

= ((tan8 x/ 8) + (tan3 x/ 3) + (tan4 x/ 4)) + c