I=π2∫02−sinx2+cosxdx=π2∫022+cosxdx−π2∫0sinx2+cosxdx* Tính J=π2∫022+cosxdx
Ta có:
2+cosx=2+2cos2x2−1=2cos2x2+1=cos2x2(2+1cos2x2)=cos2x2(tan2x2+3)
⇒J=π2∫02cos2x2(tan2x2+3)dx
Đặt t=tanx2⇒2dt=dxcos2x2
⇒J=1∫04dtt2+3
Đặt t=√3tanu⇒dt=√3(tan2u+1)du
⇒J=π6∫04√3(tan2u+1)du3tan2u+3=4√33π6∫0du=2π√39
* Tính K=π2∫0sinx2+cosxdx
Đặt v=2+cosx⇒−dv=sinxdx
⇒K=3∫2dvv=ln3−ln2=ln32
Vậy I=2π√39−ln32