Question

计算从0到π的定积分∫[x/(4+sin²x)]dx

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

利用三角恒等式和分部积分
∫x(sinx)^4dx
=
(3/8)∫xdx
-
(1/2)∫x*cos(2x)dx
+
(1/8)∫x*cos(4x)dx
=
(3/16)x^2
-
(1/2)*(1/2)[x*sin(2x)-∫sins(2x)dx]
+
(1/8)*(1/4)[x*sin(4x)-∫sin(4x)dx]
=
(3/16)*x^2
-
(1/4)x*sin(2x)-(1/8)*cos(2x)
+
(1/32)x*sin(4x)+(1/123)*cos(4x)
+
c
于是
y=∫x(sinx)∧4dx
对x定积分，x从0到π
=
(3/16)*x^2
-
(1/4)x*sin(2x)-(1/8)*cos(2x)
+
(1/32)x*sin(4x)+(1/123)*cos(4x)
|_0^π
=
(3/16)*π^2