Question

求一个式子不定积分：分子是1,分母是〔3+(cosx)平方〕

Answer 1

原式

=∫(sin^2x+cos^2x)/(3sin^2x+4cos^2x)dx

=∫(tan^2x+1)/(3tan^2x+4)dx

=∫sec^2x/(3tan^2x+4)dx

=∫1/(3tan^2x+4)dtanx

=1/3∫1/(tan^2x+4/3)dtanx

=1/3*1/(2/√3)arctan[tanx/(2/√3)]+C

=1/(2√3)arctan[tanx/(2/√3)]+C