Question

积分问题。

积分问题。高数，求下面定积分

Answer 1

设t=cosu，则
dt=-sinudu
t=0时，u=π/2
t=1时，u=0
原式=-∫(π/2→0)u·√(1-cosu)·sinudu
=∫(0→π/2)u·√2·sin(u/2)·sinudu
=√2/2·∫(0→π/2)u·[cos(u/2)-cos(3u/2)]du
【此处应用积化和差公式】
=√2·∫(0→π/2)u·d[sin(u/2)-1/3·sin(3u/2)]
=√2·u·[sin(u/2)-1/3·sin(3u/2)] |(0→π/2)
-√2·∫(0→π/2)[sin(u/2)-1/3·sin(3u/2)]du
=π/3+2√2·[cos(u/2)-1/9·cos(3u/2)] |(0→π/2)
=π/3+20/9

追答

还没写完就发出去了，
最后答案是
π/2+20/9-16/9·√2

最后答案是
π/3+20/9-16/9·√2

Answer 2

设x＝arccost
t＝cosx∈[0,1]
x∈[π/2,0]
J＝∫x·√(1-cosx)d cosx
＝∫x·sinx·√2·sin(x/2) dx
＝4∫x/2·2sin(x/2)·cos(x/2)·√2·sin(x/2)d(x/2)
＝8√2∫m·(sinm)^2d(sinm)
＝8√2 /3 ∫m d(sinm)^3
＝8√2 /3 m·(sinm)^3-8√2 /3∫(sinm)^3dm

后面的积分化成
∫(1-(cosm)^2)·sinm dm
＝-∫(---------)d cosm
最后代入全部积分区间
注意m∈(π/4,0)
从π/4积分到0