求sin^8x的不定积分

Question

求sin^8x的不定积分

Answer 1

∫(sinx)^8dx=∫((sinx)^2)^4dx
=∫((1-cos(2x))/2)^4dx (应用倍角公式)
=∫(1-4cos(2x)+6(cos(2x))^2-4(cos(2x))^3+(cos(2x))^4)dx
=∫(1-4cos(2x)+6(cos(2x))^2+(cos(2x))^4)dx-4∫(cos(2x))^3dx
=∫(1-4cos(2x)+3(1+cos(4x))+((1+cos(4x))/2)^2)dx-4∫(cos(2x))^2*cos(2x)dx
(应用倍角公式)
=∫(35/8-4cos(2x)+7cos(4x)/2+cos(8x)/8)dx-2∫(1-(sin(2x))^2)d(sin(2x))
(应用倍角公式)
=35x/8-2sin(2x)+7sin(4x)/8+sin(8x)/64-2(sin(2x)-(sin(2x))^3/3)+C (C是常数)
=35x/8-4sin(2x)+7sin(4x)/8+sin(8x)/64+2(sin(2x))^3/3+C.