#HLWRC高数#求解不定积分∫cos²xdx/(sinx+(√3)cosx)和∫sin²x/(3cosx+4sinx+5)dx？

Question

#HLWRC高数#求解不定积分∫cos²xdx/(sinx+(√3)cosx)和∫sin²x/(3cosx+4sinx+5)dx，勾股定理。

Answer 1

朋友，您好！完整详细清晰过程rt所示，此题目运算量比较大，望能详细看，希望能帮到你解决问题

Answer 2

（1）令A=∫(cosx)^2dx/(sinx+√3*cosx)，B=∫(sinx)^2dx/(sinx+√3*cosx)
3A-B=∫[3(cosx)^2-(sinx)^2]/(sinx+√3*cosx)dx
=∫(√3*cosx-sinx)dx
=√3*sinx+cosx+C1，其中C1是任意常数
A+B=∫1/(sinx+√3*cosx)dx
=(1/2)*∫1/sin(x+π/3)dx
=(1/2)*∫csc(x+π/3)d(x+π/3)
=(1/2)*ln|csc(x+π/3)-cot(x+π/3)|+C2，其中C2是任意常数
两式相加，4A=√3*sinx+cosx+(1/2)*ln|csc(x+π/3)-cot(x+π/3)|+C
原式=A=(1/2)*sin(x+π/6)+(1/8)*ln|csc(x+π/3)-cot(x+π/3)|+C，其中C是任意常数
（2）因为3cosx+4sinx+5
=5cos[x-arccos(3/5)]+5
=10{cos{[x-arccos(3/5)]/2}}^2
令t=[x-arccos(3/5)]/2，则x=2t+arccos(3/5)，dx=2dt
原式=∫{sin[2t+arccos(3/5)]}^2/10(cost)^2*2dt
=(1/5)*∫[(3/5)*sin2t+(4/5)*cos2t]^2/(cost)^2dt
=(1/5)*∫[(9/25)*(sin2t)^2+(16/25)*(cos2t)^2+(24/25)*sin2tcos2t]/(cost)^2dt
=(1/125)*∫{9*(2sintcost)^2+16*[2(cost)^2-1]^2+48*sintcost[2(cost)^2-1]}/(cost)^2dt
=(1/125)*∫[36*(sint)^2*(cost)^2+64*(cost)^4-64*(cost)^2+16+96*sint(cost)^3-48*sintcost]/(cost)^2dt
=(1/125)*∫[36*(sint)^2+64*(cost)^2-64+16*(sect)^2+48*sin2t-48*tant]dt
=(1/125)*∫[16*(sect)^2+48*sin2t-48*tant-28*(sint)^2]dt
=(1/125)*∫[16*(sect)^2+48*sin2t-48*tant-14+14*cos2t)]dt
=(1/125)*(16*tant-24*cos2t-48*ln|sect|-14t+7*sin2t)+C
=(1/125)*{16*tan{[x-arccos(3/5)]/2}-24*cos[x-arccos(3/5)]-48*ln|sec{[x-arccos(3/5)]/2}|-7x+7*arccos(3/5)+7*sin[x-arccos(3/5)]}+C
其中C是任意常数