Tính tích phân:
I=π2∫0√1−√3sin2x+2cos2xdx
Bài giải:
Ta có:
1−√3sin2x+2cos2x
=sin2x−2√3sinxcosx+3cos2x
=(sinx−√3cosx)2
=[2(12sinx−√32cosx)]2
=[2sin(x−π3)]2
Suy ra:
√1−√3sin2x+2cos2x
=√[2sin(x−π3)]2
=2|sin(x−π3)|
Do đó:
I=π2∫0√1−√3sin2x+2cos2xdx
=−2π3∫0sin(x−π3)dx+2π2∫π3sin(x−π3)dx
=2cos(x−π3)|π30−2cos(x−π3)|π2π3
=2(1−12)−2(√32−1)
=3−√3