∫cos(2x+1)^5dx=?

5/16 *sin(2x+1)+5/96 *sin[3(2x+1)]+1/160 *sin[5(2x+1)]

∫[cos(2x+1)]^5dx
= (1/2)∫ [cos(2x+1)]^4 d sin(2x+1)
=(1/2)∫ {1-2(sin(2x+1))^2+ [sin(2x+1)]^4} d sin(2x+1)
=(1/2)[ sin(2x+1) - 2(sin(2x+1))^3/3 + [sin(2x+1)]^5/5 ] +C

给一个通用解法，可以对任意的∫cos^mxdx来解
首先给一个小公式：
∫cos^m(x)dx=∫cos^(m-1)(x)d(sinx)=sinxcos^(m-1)x+∫(m-1)sin^2xcos^(m-2)xdx
=sinxcos^(m-1)x+∫(m-1)(1-cos^2x)cos^(m-2)xdx
=sinxcos^(m-1)x+∫(m-1)cos^(m-2...