∫sin^k x dx 积分上下限是0和π 时怎么做?
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先求∫(0->x)sin(x-t)^2dt
=∫(0->x)(1-cos(2x-2t)/2 dt
=1/2∫(0->x)dt-1/2∫(0->x)cos(2x-2t)dt
=x/2+1/4∫(0->x)cos(2x-2t)d(2x-2t)
=x/2+1/4sin(2x-2t)|(0->x)
=x/2+1/4(sin(2x-2x)-sin(2x-2*0)
=x/2+sin2x/4
所以
d/dx∫(0->x)sin(x-t)^2dt
=d(x/2+sin2x/4)/dx
=1/2+1/4*cos2x*2
=1/2+cos2x /2
∫(上限л/2下限0)sin^2(x/2)dx
∫(上限л/2下限0)(1-cosx)/2 dx
=(x-sinx)/2(上限л/2下限0)
=(л/2-1)/2
=л/4-1/2
所以
∫sin^k x dx 积分上下限分别是π和0=2×∫sin^k x dx 积分上下限分别是π/2和0这个时候就要根据公式来做了,你自己再搜一下就知道了,要分k是奇数和偶数两种情况
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