∫(√sinx)cos∧5(x)dx
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∫(√sinx)cos∧5(x)dx
∫(√sinx)cos∧5(x)dx
∫(√sinx)cos∧5(x)dx
原式=∫[(sinx)^(1/2)][(cosx)^4]d(sinx)
=(2/3)∫[(cosx)^4]d[(sinx)^(3/2)]
令(sinx)^(3/2)=u,则:sinx=u^(2/3), ∴(sinx)^2=u^(4/3),
∴(cosx)^4=[1-(sinx)^2]^2=[1-u^(4/3)]^2=1-2u^(4/3)+u^(8/3).
∴原式=(2/3)∫[1-2u^(4/3)+u^(8/3)]du
=(2/3)∫du-(4/3)∫u^(4/3)du+(2/3)∫u^(8/3)du
=(2/3)u-(4/3)×(3/7)u^(7/3)+(2/3)×(3/11)u^(11/3)+C
=(2/3)(sinx)^(3/2)-(4/7)[(sinx)^(3/2)]^(7/3)
+(2/11)[(sinx)^(3/2)]^(11/3)+C
=(2/3)sinx√sinx-(4/7)(sinx)^(7/2)+(2/11)(sinx)^(11/2)+C
=(2/3)sinx√sinx-(4/7)(sinx)^3√sinx+(2/11)(sinx)^5√sinx+C.
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