- ∫(0->π/2) (1+cosx)²sin³x(1+2cosx)dx
问题描述：
- ∫(0->π/2) (1+cosx)²sin³x(1+2cosx)dx
答
∫(0->π/2) (1+cosx)²sin³x(1+2cosx)dx
=∫(0->π/2) (1+2cosx+cos^2(x))sin³x(1+2cosx)dx
=∫(0->π/2) (1+2cosx+cos^2(x)+2cosx+4cos^2(x)+2cos^3(x))sin³xdx
=∫(0->π/2) [1+4cosx+5cos^2(x)+2cos^3(x)]sin³xdx
=∫(0->π/2) sin³xdx+4∫(0->π/2)cosxsin³xdx+5∫(0->π/2)cos^2(x)sin³xdx+2∫(0->π/2)cos^3(x)sin³xdx
=2/3+sin^4(x)(0,π/2)+5∫(0->π/2)sin^3(x)dx-5∫(0->π/2)sin^5(x)dx+1/8∫(0->π/2)sin^3(2x)d(2x)
=2/3+1+5*2/3-5*4*2/(5*3)+1/6
=2.5