Question

计算定积分∫(π/2 ~ -π/2) x（sinx+cosx）^2 dx 函数y=sin^4x+cos^4x的导数是

Answer 1

∫x(sinx+cosx)²
dx
=∫x(1+sin2x)
dx
=∫x
dx+∫xsin2x
dx
=1/2*x²-1/2*∫x
d(cos2x)
=1/2*x²-1/2*xcos2x+1/2*∫cos2x
dx
=1/2*x²-1/2*xcos2x+1/4*∫cos2x
d(2x)
=1/2*x²-1/2*xcos2x+1/4*sin2x
然后将积分上下限带进去计算即可
y=sin^4x+cos^4x
y'=4sin³x*(sinx)'+4cos³x*(cosx)'
=4sin³x*cosx-4cos³x*sinx
=4sinxcosx(sin²x-cos²x)
=2sin2x*(-cos2x)
=-2sin2x*cos2x
=-cos4x

Answer 2

将(1)式两边平方可得：(sinx+cosx)^2=2，化简得：sinxcosx=1/2
sin^4x+cos^4x
=(sin^2x+cos^2x)^2
-2sin^2xcos^2x
=1-2*(1/2)^2
=1/2