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](/uploads/image/z/2788831-55-1.jpg?t=%E8%AE%A1%E7%AE%97%E4%BB%8E0%E5%88%B0%CF%80%E7%9A%84%E5%AE%9A%E7%A7%AF%E5%88%86%E2%88%AB%5Bx%2F%284%2Bsin%26sup2%3Bx%29%5Ddx%E5%8F%AF%E7%94%A8%E5%85%AC%E5%BC%8F%E2%88%AB%28%E4%B8%8A%E9%99%90a%2C%E4%B8%8B%E9%99%900%29f%28x%29dx%3D%E2%88%AB%28%E4%B8%8A%E9%99%90a%2C%E4%B8%8B%E9%99%900%29f%28a-x%29dx%E7%AD%94%E6%A1%88%E4%B8%BA%CF%80%26sup2%3B%2F%284%E2%88%9A5%29%2C%E5%85%88%E7%AE%97%E5%87%BA%E5%8E%9F%E5%87%BD%E6%95%B0%E8%BF%98%E6%98%AF%E8%AE%A1%E7%AE%97%E4%B8%8D%E4%BA%86%E5%8E%9F%E5%87%BD%E6%95%B0%E6%98%AF1%2F%282%E2%88%9A5%29%26%238226%3B%5Barctan%28%E2%88%9A5%2F2%26%238226%3Btanx%29+%2B+C%E4%BB%A3%E5%85%A5x%3D%CF%80%E5%90%8E)
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)