∫(cosx)^4/(sinx)^3dx.不是x^4,是三角函数的次方喔!

原式=-∫(cosx)^4d(cosx)/(1-cos²x)²=-∫{[1/(1-y)²+1/(1+y)²-3/(1-y)-3/(1+y)]/4+1}dy (令cosx=y)=[3ln│(1+y)/(1-y)│-2y/(1-y²)]/4-y+C=[3ln│(1+cosx)/(1-cosx)│-2cosx/sin²x)]/...