Question

∫(cosx∧5*sin2x)dx上限是π/2下限是0的积分怎么求

Answer 1

显然(cosx)^2=1/2cos2x+1/2 所以∫ (cosx)^4 dx =∫ 1/4(cos2x+1)^2 dx =∫ 1/4 *(cos2x)^2 +1/2 *cos2x +1/4 dx =∫ 1/8 *cos4x +1/2 *cos2x +3/8 dx = -1/32 *sin4x -1/4 *sin2x +3x/8 代入上下限π/2和0 得到积分值=3π/16

Answer 2

∫(cos⁵x·sin2x)dx

=2∫(cos⁵x·sinxcosx)dx

=2∫(cos⁶x)(sinxdx)

=-2cos⁶xd(cosx)

=(-2/7)cos⁷x+C

∴∫(0,π/2)(cos⁵x·sin2x)dx

=(-2/7)cos⁷(π/2)-(-2/7)cos⁷(0)=2/7