计算从0到π的定积分∫[x/(4+sin²x)]dx可用公式∫(上限a,下限0)f(x)dx=∫(上限a,下限0)f(a-x)dx答案为π²/(4√5),先算出原函数还是计算不了原函数是1/(2√5)•[arctan(√5/2•tanx) + C代入x=π后
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计算从0到π的定积分∫[x/(4+sin²x)]dx可用公式∫(上限a,下限0)f(x)dx=∫(上限a,下限0)f(a-x)dx答案为π²/(4√5),先算出原函数还是计算不了原函数是1/(2√5)•[arctan(√5/2•tanx) + C代入x=π后
计算从0到π的定积分∫[x/(4+sin²x)]dx
可用公式∫(上限a,下限0)f(x)dx=∫(上限a,下限0)f(a-x)dx
答案为π²/(4√5),
先算出原函数还是计算不了
原函数是1/(2√5)•[arctan(√5/2•tanx) + C
代入x=π后,tan(π)=0,tan(0)=0,那结果岂不是=0?
计算从0到π的定积分∫[x/(4+sin²x)]dx可用公式∫(上限a,下限0)f(x)dx=∫(上限a,下限0)f(a-x)dx答案为π²/(4√5),先算出原函数还是计算不了原函数是1/(2√5)•[arctan(√5/2•tanx) + C代入x=π后
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)