万能公式 ∫R(sinx, cosx)dx
= ∫R[2u/(1+u^2), (1-u^2)/(1+u^2)]2du/(1+u^2)
凑幂公式
∫f(x^n)x^(n-1)dx = (1/n)∫f(x^n)dx^n
∫[f(x^n)/x]dx = (1/n)∫[f(x^n)/x^n]dx^n
∫(asinx+bcosx)dx/(psinx+qcosx)型,
设 asinx+bcosx = A(psinx+qcosx) + B(psinx+qcosx)'
降幂递推公式
I
I
I
∫(cosx)^2/(sinx)^3*dx
=∫cosx/(sinx)^3*d(sinx)
=-2∫cosxd[1/(sinx)^2]
=-2cosx/(sinx)^2+2∫1/(sinx)^2*d(cosx)
=-2cosx/(sinx)^2+2∫1/[1-(cosx)^2]*d(cosx)
=-2cosx/(sinx)^2+2*(1/2)ln|(1+cosx)/(1-cosx)|+C
=-2cosx/(sinx)^2+ln[(1+cosx)/(1-cosx)]+C
什么?