Question

高数 ∫1+sin²2X/1+cos²2X

Answer 1

I = ∫[(1+sin²2x)/(1+cos²2x)]dx = ∫[(3-cos4x)/(3+cos4x)]dx
= ∫[(6-3-cos4x)/(3+cos4x)]dx = (3/2)∫[1/(3+cos4x)]d(4x) - ∫dx
令 tan2x = u, 则 cos4x = (1-u^2)/(1+u^2), d(4x) = 2du/(1+u^2)
I1 = ∫[1/(3+cos4x)]d(4x) = ∫du/(2+u^2) = (1/√2)arctan(u/√2)
= (√2/2)arctan(tan2x/√2)
I = (3√2/4)arctan(tan2x/√2) - x + C

Answer 2

∫(1+sin²2X)/(1+cos²2X)dX
=∫(3-1-cos²2X)/(sin²2X+2cos²2X)dX
=∫3/(sin²2X+2cos²2X)dX-∫dX
=3∫sec²2X/(tan²2X+2)dX-X
=(3/2)∫1/(tan²2X+2)dtan²2X-X
=(3/(2√2))arctan(tan((2X)/√2))-X+C