求不定积分i1=sinx/(acosx+bsinx)i2=cosx/(acosx+bsinx)求i1i2求不定积分i1=sinx/(acosx+bsinx)i2=cosx/(acosx+bsinx)求i1i2

Question

求不定积分i1=sinx/(acosx+bsinx) i2=cosx/(acosx+bsinx)求i1 i2
求不定积分i1=sinx/(acosx+bsinx) i2=cosx/(acosx+bsinx)求i1 i2

Answer 1

这两个的做法是一样的,以sinx/(acosx+bsinx)为例说明.
思路：将sinx表示为acosx+bsinx及其导数的线性组合
sinx＝b/(a^2+b^2)×（acosx+bsinx）－a/(a^2+b^2)×（－asinx＋bcosx）＝b/(a^2+b^2)×（acosx+bsinx）－a/(a^2+b^2)×（acosx+bsinx）'
所以,∫sinx/(acosx+bsinx)dx＝bx/(a^2+b^2)－a/(a^2+b^2)×ln|acosx+bsinx|＋C
同理,∫cosx/(acosx+bsinx)dx＝ax/(a^2+b^2)＋b/(a^2+b^2)×ln|acosx+bsinx|＋C