沈浩德
春芽
共回答了20个问题采纳率:80% 举报
二楼做得有一点问题 设T=∫(0,π)[x/(4+sin²x)]dx T=∫(π,0)[(π-x)/(4+sin²(π-x)]d(π-x) (用π-x代换x) ==>T=-∫(π,0)[(π-x)/(4+sin²x)]dx ==>T=∫(0,π)[(π-x)/(4+sin²x)]dx ==>T=π∫(0,π)[1/(4+sin²x)]dx-∫(0,π)[x/(4+sin²x)]dx ==>T=π∫(0,π)[1/(4+sin²x)]dx-T ==>2T=π∫(0,π)[1/(4+sin²x)]dx T=(π/2)∫(0,π)[1/(4+sin²x)]dx 下面拆为两个区间,否则会有瑕点 =(π/2)∫(0,π/2)[sec²x/(4sec²x+tan²x)]dx+(π/2)∫(π/2,π)[sec²x/(4sec²x+tan²x)]dx =(π/2)∫(0,π/2)[1/(4sec²x+tan²x)]d(tanx)+(π/2)∫(π/2,π)[1/(4sec²x+tan²x)]d(tanx) =(π/2)∫(0,π/2)[1/(4+5tan²x)]d(tanx)+(π/2)∫(π/2,π)[1/(4+5tan²x)]d(tanx) =(π/2)*1/(2√5)•arctan(√5/2•tanx) [0-->π/2]+(π/2)*1/(2√5)•arctan(√5/2•tanx) [π/2-->π] =(π/2)*1/(2√5)•π/2-(π/2)*1/(2√5)•(-π/2) =π²/(4√5)
1年前
2