1 (sinxcosx)的原函数-搜答案-智慧作业帮

0.5sinx^2

y=√2/2*sin2x+(1+cos2x)/2-1/2=√2/2*sin2x+1/2*cos2x=√3/2*sin(2x+z)其中tanz=(1/2)/(√2/2)=√2/2所以T=2π/2=π

再问：做对了，但是有一点看不懂。1/cosx+1/sinx接下来的那一步…能解释一下吗再答：因为分子是1嘛，而sinx的平方加cosx的平方就是1.所以就把1换掉再答：把分子的1换掉后，就可以分离，然

设sinx+cosx=t属于[-√2,√2]=>t^2=1+2sinxcosx=〉sinxcosx=(t^2-1)/2f(x）=（t^2-1）/2(1+t)=(t-1)/2属于[-(√2+1)/2,(

y=√3sin2x-cos2x=2sin(2x-30°)ymin=-2

f(x)=(1-cos2x)/2+(√3/2)sin2x+1/2=(√3/2)sin2x-(1/2)cos2x+1=sin(2x-π/6)+1（1）最小正周期T=2π/2=π（2）f(x)max=2,

f(x)=√3sinxcosx+cos?x=√3/2sin2x+(cos2x+1)/2=√3/2sin2x+1/2cos2x+1/2=sin(2x+π/6)+1/2f(x)的最小正周期为2π/2=π∵

-1/4cos2x+c

1.求函数y=sinx-cosx+sinxcosxx∈（0,π）的最大值最小值设t=sinx–cosx所以t²=1–2sinxcosx,则sinxcosx=1-t²/2因为t=si

答：设t=tanx,sinx=tcosx代入(sinx)^2+(cosx)^2=1有：(t^2+1)(cosx)^2=1(cosx)^2=1/(1+t^2)所以：f(tanx)=(sinx)^2+si

令sinx-cosx=a则a^2=(sinx)^2+(cosx)^2-2sinxcosx=1-2sinxcosx所以sinxcosx=(1-a^2)/2所以y=1+a+(1-a^2)/2=-a^2/2

（1）f(x)=（cosx）^2+√3sinxcosx-1/2=1/2(1+cos2x)+√3/2sinxcosx-1/2=1/2cos2x+√3/2sin2x=sin(2x+30°),f(x)最大值

x=0代入f(0)=cos0-sin0+2(3sin0cos0+1)=1-0+2(0+1)=3

y=sinxcosx-1=1/2+sinxcosx-3/2=(1+2sinxcosx)/2-3/2=(sinx+cosx)^2/2-3/2=sin^2(x+π/4)-3/2所以最大值是1-3/2=-1

g

sinx+cosx=t√2sin(x+∏/4)=t-√2≤t≤√21+2sinxcosx=t²sinxcosx=(t²-1)/2y=1+sinx+cosx+sinxcosx=1+t

=sinxcosx/1+sinxcosx/sin^2x+sinxcosx/cos^2x=sinxcosx/1+sinxcosx/1=sinxcosx/2=2sinxcosx/4=sin2x/4sin2

∫1/(sinxcosx)^3dx=8∫1/(sin2x)^3dx=-4∫1/(sin2x)^4dcos2x=-4∫1/[1-（cos2x)^2]^2dcos2x设cos2x=y上式=-4∫1/[1-

y=1+2sinxcosx+sinx+cosx=sin²x+cos²x+2sinxcosx+sinx+cosx=(sinx+cosx)²+sinx+cosx=(sinx+