sint-cost

∫sint/(cost+sint)dt∫sint/(cost+sint)dt=(1/2)∫[(sint+cost)+(sint-cost)]/(cost+sint)dt=(1/2)∫dt+(1/2)∫(sint-cost)/(cost+si

∫cost/(sint+cost)dt

sint/(1-cost)求导f'(t)=[(sint)'(1-cost)-(sint)(1-cost)']/(1-cost)=[cost(1-cost)-sintsint]/(1-cost)=[cost-(cost+sint)]/(1-c

∫dt/(1+sint+cost)

(sintcost)^2的不定积分∫(sint·cost)²dt=∫(½·sin2t)²dt=1/4·∫(sin2t)²dt=1/4·∫(1-cos4t)/2dt=1/8·∫(1-cos4t)dt=1

sint/1-cost=什么sint=2sin二分之t乘以cos二分之t1-cost=2（sin二分之t）方二者一比=cot二分之t

为什么∫(2sint+cost)+2(2cost-sint)/2sint+costdt=(t+2ln|2sint+cost|)+C?因为∫[(2sint+cost)+2(2cost-sint)]/(2sint+cost)dt=∫(2sint

(cost-sint)/(sint-cost)的结果不知道等不等于-(sint-cost)/(sint-cost)也就是-1它们是相等的

求fxo(sint+cost·sint)dt的最大值求fxo(sint+cost·sint)dt的最大值答：(0→x)∫(sint+costsint)dt=(0→x)-∫(1+cost)d(cost)=(0→x)-cost-(1/2)(co

如何直接看出0到pai/2定积分cost/(sint+cost)与sint/(sint+cost)相等?只需令x=pi/2-t,则当x=0,t=pi/2,当x=pi/2,t=0,dx=-dt,那么∫(0,pi/2)cosx/(sinx+co

cost的平方/sint的不定积分cost^2/sintdt=(1-sint^2)/sintdt=1/sintdx-sintdt=csctdt-sintdt=lnlcsct-cottl+cost+c

(1/(1+sint+cost))dt不定积分

sint/(1-cost)=cost/2怎么推?

t趋于0时为什么lim(sint+cost-1)/t=lim(cost-sint)=1)你们没学过吗0/0型求极限,分子分母同时对t求导数,再把t=0带进导数里.

x=sint-costy=sint+cost求它得普通方程x=sint-costy=sint+cost则：x+y=2sintx-y=-2cost所以：(x+y)^2+(x-y)^2=2X平方+Y平方=2

(sint+cost)/(cost-sint)怎么算的,等于tan(t+π/4)(sint+cost)/(cost-sint)=[(sint+cost)/√2]/[(cost-sint)/√2]=(sintcosπ/4+costsinπ/4

∫sint^2cost^2dt=?怎么算啊∫sint^2cost^2dt=1/4∫（sin2t)^2dt=1/4∫(1-cos4t)/2dt=1/8∫dt-1/8∫cos4tdt=1/8t-1/32sin4t+c

∫sint*sinwtdt怎么求积分?还有∫cost*coswtdt?用积化和差公式：sinxsiny=(1/2)[cos[(x-y)-cos(x+y)],cosxcosy=(1/2)[cos(x+y)+cos(x-y)]∫sint*sin

1/(sint)^2和1/(cost)^2的不定积分∫[1/(sint)^2]dt=-∫dcott=-cott+C∫[1/(cost)^2]dt=∫dtant=tant+C上面这两个属于基本公式,最好记住,对做题有好处.

(sint)^9*(cost)^3的不定积分怎么求原式＝∫［（sint）^9］［（cost）^2］d（sint）　　＝∫［（sint）^9］［1－（sint）^2］d（sint）　　＝∫［（sint）^9］d（sint）－∫［（sint）^