∫(x (sinx cosx))

f(x)=sin²x+sinxcosx=[1-cos(2x)]/2+sin(2x)/2=sin(2x)/2-cos(2x)/2+1/2=(√2/2)sin(2x-π/4)+1/2最小正周期T

sin²x-sinxcosx+2cos²x=1-sinxcosx+cos²x=1-½sin2x+(1+cos2x)/2=3/2+½cos2x-

sin^2xtanx+cos^2x/tanx+2sinxcosx-(1+cosx/sinxcosx)=sin^3x/cosx+cos^3x/sinx+2sinxcosx-(1+cosx/sinxcos

=§sinx/(1+sin^4x)dsinx设sinx=t原式=1/根号8§1/(t^2-根号2t+1)-1/(t^2+根号2t+1)dt然后就是代公式了!令x=2sint则原式=1/4§1/sin^

第一题很简单啊∫(sinXcosx)/(1+sin^4X)dx=0.5∫1/(1+sin^4X)d（sin^2x）把sin^2x看成整体会了吧第二题很简单啊∫dx/(X^2(4-X^2)^0.5换元啊

2*(sinxcosx-cos平方x)+1=2sinxcosx-2cos²x+1=2sinxcosx-cos2x=sin2x-cos2x=√2sin(2x-45°)

f(x)=cos^2x+(1/2)sin2x.f(x)=(1+cos2x)/2+(1/2)sin2x.=(1/2)(sin2x+cos2x)+1/2.∴f(x)=(1/2)√2sin(2x+π/4)+

=(1+cos2x)/2+sin2x/2=根号2/2sin(2x+pi/4)+1/2

sin²x=(1-cos2x)/2sinxcosx=1/2*sin2xsin²X-sinXcosX=1/2-(cos2x+sin2x)/2=1/2-√2/2*sin(2x+45°)

怎么感觉cosx应该是平方啊再问：嗯的，打错了再答：(cosx)^2=(1+cos2x)/2sinxcosx=1/2*sin2x所以原式=(1+cos2x)/2-根号3/2*sin2x+1=1/2*c

1f（x）=2√3sinxcosx+2cos2x-1=√3sin2x+cos2x=2sin(2x+π/6)最小正周期T=2π/2=π∵x∈[0,π/2]∴2x+π/6∈[π/6,7π/6]∴2x+π/

f(x)=2sinxcosx+cos2x=sin2x+cos2x=√2sin(2x+π/4)2x+π/4=π/2+2kπ时f(x)有最大值f(x)=√2x=π/8+kπ2x+π/4=3π/2+2kπ时

再答：您好，很高兴能回答您的问题，希望对您有帮助！答案见上图。很高兴为你解答，仍有不懂请追问，满意请采纳，谢谢！----【百度懂你】团队提供

f(x)=sin2x+cos2x=√2sin(2x+π/4)所以T=2π/2=π最大值=√2f(θ+π/8)=√2sin(2θ+π/4+π/4)=√2cos2θ=√2/3cos2θ=1/3θ锐角则si

原式等于（3sinXcosX+cos²x-sin²x）/（sin²x+cos²x）再同时除以cos²x就行了

(1)∫[(sinxcosx)/(1+sin²x)]dx,d(1+sin²x)=(2sinxcosx)dx=∫[(sinxcosx)/(1+sin²x)*1/(2sinx

∫sin2xdx/(sinx+cosx)=∫cos(π/2-2x)dx/[√2cos(π/4-x)]=√2∫cos(π/4-x)dx-(1/√2)∫dx/cos(π/4-x)=√2sin(x-π/4)

原式=sinxcosx+cos²x+(sin²x+cos²x)=1/2*sin2x+(1+cos2x)/2+1=1/2(sin2x+cos2x)+3/2=√2/2*sin

∫sinxcosx/(1+sin^4x)dx=∫sinx/(1+sin^4x)d(sinx)=1/2*∫1/(1+(sin^2x)^2)d(sin^2x)=1/2*arctan(sin^2x)+C