∫(sinx)^4(cosx)^3dx

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∫(sinx)^4(cosx)^3dx
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∫(sinx)^4(cosx)^3dx
∫(sinx)^4(cosx)^3dx

∫(sinx)^4(cosx)^3dx
∫(sinx)^4(cosx)^3dx
=∫(sinx)^4(cosx)^2d(sinx)
设t=sinx,(cosx)^2=1-(sinx)^2=1-t^2
原式=∫t^4(1-t^2)dt
=∫(t^4-t^6)dx
=(t^5)/5-(t^7)/7+C
=(1/5)*(sinx)^5-(1/7)*(sinx)^7+C