∫(- π/2→π/2) (sin⁴x + cos⁵x) dx
= 2∫(0→π/2) (sin⁴x + cos⁵x) dx
= 2 * 3!!/4!! * π/2 + 2 * 4!!/5!!
= 2 * (3 * 1)/(4 * 2 * 1) * π/2 + 2 * (4 * 2 * 1)/(5 * 3 * 1)
= 16/15 + 3π/8
定积分就是好在有这性质,省掉许多功夫,为什么这方法就不多人认识呢
如果先求
原函数的话,将会费力得多
sin⁴x = (sin²x)² = [(1 - cos2x)/2]²
= (1/4)(1 - 2cos2x + cos²2x)
= 1/4 - (1/2)cos2x + (1/4)[(1 + cos4x)/2]
= 3/8 - (1/2)cos2x + (1/8)cos4x
cos⁵x dx
= cos⁴x d(sinx)
= (cos²x)² d(sinx)
= (1 - sin²x)² d(sinx)
= (1 - 2sin²x + sin⁴x) d(sinx)
这相当于(1 - 2u² + u⁴) du,求法很简单
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