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已知函数F(x)=Cos^4x%sin^4x+8sinx/2Cosx/2 (1)求单调增区间 (2)...

解答如下: 第一问 f(x)=cos^4x+2sinxcosx-sin^4x =(c

解由f(x)=cos^4x-2sinxcosx-sin^4x =cos^4x-sin^4x-2s

(1)f(x)=cos4x-2sinxcosx-sin4x=(cos2x+sin2x)(cos2x-

∫ (sin^4x)*(cos^2x) dx=1/16*x-1/64*sin4x-1/48*(sin

f(x)=cos^4x-2sinxcosx-sin^4x =(cos^2x+sin^2x)(cos

利用二倍角公式,把函数化为cos2x和sin2x形式,由于x属于[0,π/2],所以sin2x总是正

f(x) = sin⁴x+2√3sinxcosx-cos⁴x

∫xcos^4(x/2)/sin^3(x)dx的结果为-x/(8*sin^2(x/2))-cot(x

f(x)=cos^4x-sin^4x+根号3sin2x=cos^2x-sin^2x+根号3sin2x

∫sinxcosx/(1+sin^4x)dx =∫sinx/(1+sin^4x)d(sinx)

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