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∫1/(cosx(5+3cosx)）dx,指点下怎么换吧,还是令u=tanx/2?那样挺麻烦的,虽然能做出,有没有更好的方法呢
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∫1/(cosx(5+3cosx)）dx,指点下怎么换吧,还是令u=tanx/2?那样挺麻烦的,虽然能做出,有没有更好的方法呢
▼优质解答
答案和解析
int(1/(cos(x)*(5+3*cos(x))), x)
= int((-1-u^2)/(3*u^2+u^4-4), u) and int((-1-u^2)/(3*u^2+u^4-4), u)
= int(-(3/5)/(u^2+4)+1/(5*u+5)-1/(-5+5*u), u) and int(-(3/5)/(u^2+4)+1/(5*u+5)-1/(-5+5*u), u)
= int(-(3/5)/(u^2+4), u)+int(1/(5*u+5), u)+int(-1/(-5+5*u), u) and int(-(3/5)/(u^2+4), u)+int(1/(5*u+5), u)+int(-1/(-5+5*u), u)
= -(3/5)*(int(1/(u^2+4), u))+int(1/(5*u+5), u)+int(-1/(-5+5*u), u) and -(3/5)*(int(1/(u^2+4), u))+int(1/(5*u+5), u)+int(-1/(-5+5*u), u)
= -(3/5)*(int(1/2, u1))+int(1/(5*u+5), u)+int(-1/(-5+5*u), u) and -(3/5)*(int(1/2, u1))+int(1/(5*u+5), u)+int(-1/(-5+5*u), u)
= -(3/10)*u1+int(1/(5*u+5), u)+int(-1/(-5+5*u), u) and -(3/10)*u1+int(1/(5*u+5), u)+int(-1/(-5+5*u), u)
= -(3/10)*arctan((1/2)*u)+int(1/(5*u+5), u)+int(-1/(-5+5*u), u) and -(3/10)*arctan((1/2)*u)+int(1/(5*u+5), u)+int(-1/(-5+5*u), u)
= -(3/10)*arctan((1/2)*u)+(1/5)*(int(1/(u+1), u))+int(-1/(-5+5*u), u) and -(3/10)*arctan((1/2)*u)+(1/5)*(int(1/(u+1), u))+int(-1/(-5+5*u), u)
= -(3/10)*arctan((1/2)*u)+(1/5)*(int(1/u1, u1))+int(-1/(-5+5*u), u) and -(3/10)*arctan((1/2)*u)+(1/5)*(int(1/u1, u1))+int(-1/(-5+5*u), u)
= -(3/10)*arctan((1/2)*u)+(1/5)*ln(u1)+int(-1/(-5+5*u), u) and -(3/10)*arctan((1/2)*u)+(1/5)*ln(u1)+int(-1/(-5+5*u), u)
= -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)+int(-1/(-5+5*u), u) and -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)+int(-1/(-5+5*u), u)
= -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*(int(1/(-1+u), u)) and -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*(int(1/(-1+u), u))
= -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*(int(1/u1, u1)) and -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*(int(1/u1, u1))
= -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*ln(u1) and -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*ln(u1)
= -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*ln(-1+u) and -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*ln(-1+u)
= -(3/10)*arctan((1/2)*tan((1/2)*x))-(1/5)*ln(tan((1/2)*x)-1)+(1/5)*ln(tan((1/2)*x)+1)
希望能帮助到你~
= int((-1-u^2)/(3*u^2+u^4-4), u) and int((-1-u^2)/(3*u^2+u^4-4), u)
= int(-(3/5)/(u^2+4)+1/(5*u+5)-1/(-5+5*u), u) and int(-(3/5)/(u^2+4)+1/(5*u+5)-1/(-5+5*u), u)
= int(-(3/5)/(u^2+4), u)+int(1/(5*u+5), u)+int(-1/(-5+5*u), u) and int(-(3/5)/(u^2+4), u)+int(1/(5*u+5), u)+int(-1/(-5+5*u), u)
= -(3/5)*(int(1/(u^2+4), u))+int(1/(5*u+5), u)+int(-1/(-5+5*u), u) and -(3/5)*(int(1/(u^2+4), u))+int(1/(5*u+5), u)+int(-1/(-5+5*u), u)
= -(3/5)*(int(1/2, u1))+int(1/(5*u+5), u)+int(-1/(-5+5*u), u) and -(3/5)*(int(1/2, u1))+int(1/(5*u+5), u)+int(-1/(-5+5*u), u)
= -(3/10)*u1+int(1/(5*u+5), u)+int(-1/(-5+5*u), u) and -(3/10)*u1+int(1/(5*u+5), u)+int(-1/(-5+5*u), u)
= -(3/10)*arctan((1/2)*u)+int(1/(5*u+5), u)+int(-1/(-5+5*u), u) and -(3/10)*arctan((1/2)*u)+int(1/(5*u+5), u)+int(-1/(-5+5*u), u)
= -(3/10)*arctan((1/2)*u)+(1/5)*(int(1/(u+1), u))+int(-1/(-5+5*u), u) and -(3/10)*arctan((1/2)*u)+(1/5)*(int(1/(u+1), u))+int(-1/(-5+5*u), u)
= -(3/10)*arctan((1/2)*u)+(1/5)*(int(1/u1, u1))+int(-1/(-5+5*u), u) and -(3/10)*arctan((1/2)*u)+(1/5)*(int(1/u1, u1))+int(-1/(-5+5*u), u)
= -(3/10)*arctan((1/2)*u)+(1/5)*ln(u1)+int(-1/(-5+5*u), u) and -(3/10)*arctan((1/2)*u)+(1/5)*ln(u1)+int(-1/(-5+5*u), u)
= -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)+int(-1/(-5+5*u), u) and -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)+int(-1/(-5+5*u), u)
= -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*(int(1/(-1+u), u)) and -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*(int(1/(-1+u), u))
= -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*(int(1/u1, u1)) and -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*(int(1/u1, u1))
= -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*ln(u1) and -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*ln(u1)
= -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*ln(-1+u) and -(3/10)*arctan((1/2)*u)+(1/5)*ln(u+1)-(1/5)*ln(-1+u)
= -(3/10)*arctan((1/2)*tan((1/2)*x))-(1/5)*ln(tan((1/2)*x)-1)+(1/5)*ln(tan((1/2)*x)+1)
希望能帮助到你~
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