Question

f(x)=∫[0~x]tsin√(x^2-t^2)dt 求f(x)的导数

Answer 1

设u=√（x^2-t^2),则t=√(x^2-u^2),积分上限限对应为[x,0]
f(x)
=∫[x~0]√(x^2-u^2)*sinud√(x^2-u^2)
=∫[x~0]√(x^2-u^2)*sinu[(1/2)*(-2u)/√(x^2-u^2)]du
=-∫[x~0]]usinudu
=∫[x~0]udcosu
=ucosu[x,0]-∫[x~0]cosudu
=ucosu[x,0]+sinu[x,0]
=-xcosx-sinx

所以:
f（x）‘=xsinx-2cosx.