求定积分 ∫(0~π/2)(cosx)^5*(sinx)^2 dx
问题描述:
求定积分 ∫(0~π/2)(cosx)^5*(sinx)^2 dx
答
∫(0→π/2)(cosx)^5·(sinx)²dx
=∫(0→π/2)(cosx)^4·(sinx)²d(sinx)
=∫(0→π/2)(1-sin²x)²·(sinx)²d(sinx)
=∫(0→π/2)[(sinx)^6+2(sinx)^4+(sinx)^2]d(sinx)
=[1/3(sinx)^3-2/5(sinx)^5+1/7(sinx)^7] |(0→π/2)
=1/3-2/5+1/4
=8/105
答
∫(0→π/2)(cosx)^5·(sinx)²dx
=∫(0→π/2)(cosx)^4·(sinx)²d(sinx)
=∫(0→π/2)(1-sin²x)²·(sinx)²d(sinx)
=∫(0→π/2)[(sinx)^6+2(sinx)^4+(sinx)^2]d(sinx)
=[1/3(sinx)^3-2/5(sinx)^5+1/7(sinx)^7] |(0→π/2)
=1/3-2/5+1/4
=8/105