Question

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)，求过程。
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？

Answer 1

解：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
先求不定积分∫[1/(4+sin²x)]dx的原函数
设 t=tanx，则sin²x=t/(1+t²),dx=dt/(1+t²)
==>∫[1/(4+sin²x)]dx=∫[1/(4+5t²)]dt
=[1/(2√5)]∫[1/(1+(√5t/2)²)]d(√5t/2)
=[1/(2√5)]arctan(√5t/2)
即不定积分∫[1/(4+sin²x)]dx的原函数是[1/(2√5)]arctan(√5t/2)
故T=(π/2)∫(0,π)[1/(4+sin²x)]dx
=(π/2)*{[1/(2√5)]arctan(√5t/2)}│(0,π)
=(π/2)*[1/(2√5)](π-0)
=π²/(4√5)。

Answer 2

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)

Answer 3

不知，嘻嘻

Answer 4

可惜我身边没有教材啊，很多东西和公式都不记得了。坐等高手。