第1个回答 2019-12-26
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
=
∫(tanx)^ndx
=
(tanx)^(n-1)/(n-1)
-
I
I
=
∫(sinx)^ndx
=
-cosx(sinx)^(n-1)/n
+
(n-1)I
/n
I
=
∫(cosx)^ndx
=
sinx(cosx)^(n-1)/n
+
(n-1)I
/n