Question

hello friends please help me... find the integral..

Answer 1

2 20-05-2005 23:13

u = tanx

du/dx = sec2x

I = ∫ tan4x dx

= ∫ tan2x tan2x dx

= ∫ tan2x (sec2x - 1) dx

= ∫ tan2x sec2x dx - ∫ tan2x dx

= ∫ u2 du - ∫ (sec2x - 1) dx

= (1/3)u3 - (tanx - x) + C

= (1/3)tan3x - tanx + x + C

Answer 2

the answer is as followes