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Integrate tan^4x

Hey again

Question:

Use the identity sec^2 x=tan^2 x + 1 and the substitution tanx=u to find
INT tan^4 x .dx

please help me

step by step

this has been frying my head!!!!

Reply 1

dvs

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