求∫dxa2sin2x+b2cos2x.(a、b是不全为零的非负常数

\u6574\u5e38\u6570\u6709\u6ca1\u6709\u8d1f\u7684\u554a?

\u6709\u8d1f\u6570\uff01\uff5e\u6574\u6570\u4e0d\u662f\u81ea\u7136\u6570\uff01\uff5e

\u5e38\u6570\u662f\u6307 \u6709\u786e\u5b9a\u503c\u7684\u6570\u3002\u5982\u201c1\u201d\u201c\uff0d1\u201d\u7b49\u3002

\u6574\u5e38\u6570\u662f\u6307 \u6574\u578b\u7684\u5e38\u91cf\uff08\u5373 \u6570\u5b66\u91cc\u7684 \u6574\u6570\uff09

\u8981\u5168\u90e8\u79fb\u5230\u4e00\u8fb9\u53bb\uff0c\u901a\u5e38\u60c5\u51b5\u4e0b\uff0c\u4e8c\u6b21\u9879\u7cfb\u6570\u4e3a\u6b63\uff0c\u8fd9\u65f6\u518d\u770b\u7b26\u53f7

(I)a≠0,b=0时,
dx
a2sin2x+b2cos2x
=
dx
a2sin2x
1
a2
∫csc2xdx=?
cotx
a2
+C

(II)a=0,b≠0时,
dx
a2sin2x+b2cos2x
=
dx
b2cos2x
1
b2
∫sec2xdx=
tanx
b2
+C

(III)a≠0,b≠0时,
dx
a2sin2x+b2cos2x
=
dx
cos2x(a2tan2x+b2)
=∫
sec2xdx
b2+a2tan2x
=
1
b2
d(tanx)
1+(
a
b
tanx)2
=
1
ab
d(
a
b
tanx)
1+(
a
b
tamx)2
1
ab
arctan(
a
b
tanx)+C


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