Integral of Tan^4.sec^3x
Answers
Answered by
0
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
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
Similar questions