sint+cost
cost/sint+cost=sint/cost+sint
=-∫(pi/2,0)sint/(sint+cost)dt =∫(0,pi/2)sinx/(sinx+cosx)dx 所以 ∫(0,pi/2)cosx/(sinx+cosx)dx=∫(0,pi/2)sinx/(sinx+cosx)dx =(1/2)[∫...
证明tant=sint/1+cost
推测公式左端有误,应该是tan(t/2). 证明如图示：
sint+cost=
sinx+cosx=√2(√2/2sinx+√2/2cosx)=√2(sinxcos45+cosxsin45)=√2sin(x+45)楼主请把里面的x换成t即可~
...分母变成1/(sint+cost),然后不会做了
解：设x=sint，则 原式=∫(0,π/2)costdt/(sint+cost)。再设I=∫(0,π/2)costdt/(sint+cost)，t=π/2-y，则I=∫(0,π/2)...
高数不定积分
令x=sint，则dx=costdt 原式=∫costdt/(sint+cost)=(1/2)*∫[(sint+cost)+(cost-sint)]/(sint+cost)dt =(1/2)*∫dt+(1/2)...
,高等数学
A-B=∫(cost-sint)/(sint+cost)dt=∫d(sint+cost)/(sint+cost)=ln|sint+cost|+C 两式相加，2A=t+ln|sint+cost|+C A=t/2+(1...
多少的导数是sint+sintcost
根据题意有：y'=sint+sintcosty'=sint(1+cost)=sint[1+2cos^2(t/2)-1]=2cos^2(t/2)*2sin(t/2)cos(t/2)=4cos^3(t/2)sin(t/2)两边积分得到：∫...
这步用的是啥公式呀?
1/（sint+cost）=1/√2[1/（costcosπ/4+sintsinπ/4）=1/√2[1/cos（t-π/4）]=1/√2sec（t-π/4）
∫cost/(sint+cost)dt=
dt =(1/2)∫ dt + (1/2)∫(sint-cost)/(cost+sint)dt =(1/2)∫ dt - (1/2)∫dln(sint+cost)=(1/2)t - (1/2)ln|sint+cost| + C ...
y=sint/sint+cost的导数
y=sint/(sint+cost)y'=[cost(sint+cost)-sint(cost-sint)]/(sint+cost)^2 =(costsint+cos^2t-sintcost+sin^2)/(sint+cost)^2 =1/(sint+cost)^2 ...