(2004 · Anhui) if f (SiNx) = 2-cos2x, then f (cosx) is equal to () A. 2-sin2x B. 2+sin2x C. 2-cos2x D. 2+cos2x

(2004 · Anhui) if f (SiNx) = 2-cos2x, then f (cosx) is equal to () A. 2-sin2x B. 2+sin2x C. 2-cos2x D. 2+cos2x

∵f（sinx）=2-（1-2sin2x）=1+2sin2x，
∴f（x）=1+2x2，（-1≤x≤1）
∴f（cosx）=1+2cos2x=2+cos2x．
Therefore, D is selected

If f (SiNx) = 3-cos2x, then f (cosx)=______ .

f（cosx）=f[sin(π
2−x)]
=3-cos（π-2x）
=3+cos2x．
So the answer is: 3 + cos2x

Find the indefinite integral ∫ (cos2x) / (SiNx + cosx) DX

∫cos2x/(sinx+cosx)dx
=∫(cos²x-sin²x)/(sinx+cosx)dx
=∫(cosx-sinx)dx
=sinx+cosx+C
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The indefinite integral of cos2x \ (cosx + SiNx)? Please write down the steps to solve the problem clearly, thank you!

∫cos2x/(cosx+sinx)dx
=∫[(cosx)^2-(sinx)^2]/(cosx+sinx)dx
=∫cosx-sinx dx
=∫cosxdx-∫sinxdx
=sinx-(-cosx)+C
=sinx+cosx+C

Solving indefinite integral: ∫ cosx / (SiNx + cosx) DX

∫cosx/(sinx+cosx) dx
= (1/2)∫[(cosx+sinx)+(cosx-sinx)]/(sinx+cos)] dx
= (1/2)∫ dx + (1/2)∫(cosx-sinx)/(sinx+cosx) dx
= x/2 + (1/2)∫d(sinx+cosx)/(sinx+cosx)
= (1/2)(x+ln|sinx+cosx|) + C
reference resources:
A=∫cosx/（sinx+cosx)dx
B=∫sinx/(sinx+cosx)dx
A+B=∫(cosx+sinx)/(sinx+cosx)dx =∫dx =x+c (1)
A-B=∫(cosx-sinx)/(sinx+cosx)dx =∫(d(cosx+sinx)/(sinx+cosx)=ln(cosx+sinx)+c (2)
[1) + (2)] / 2
A=∫cosx/(sinx+cosx)dx =x/2+1/2*ln(cosx+sinx)+c

Find the indefinite integral ∫ cos2x / (SiNx) ^ 2 DX

∫cos2x/sin²xdx=∫(cos²x-sin²x)/sin²xdx=∫(cos²x+sin²x-2sin²x)/sin²xdx=∫(1-2sin²x)/sin²xdx=∫(1/sin²x-2)dx=∫1/sin²xdx-∫2dx=-cotx-2x+C...

Find the indefinite integral ∫ (cos2x) / (sin ^ 2x) (COS ^ 2x) DX For an indefinite integral, ∫ (cos2x) / (sin ^ 2x) (COS ^ 2x) DX, ∫ cos double x divided by sin square x times cos square x DX

∫cos(2x)dx/(sin²xcos²x)=∫4cos(2x)dx/sin²(2x)
=2∫d(sin(2x))/sin²(2x)
=-2 / sin (2x) + C (C is the integral constant)

∫ (cos2x / sin ^ 2x) DX needs a process

Cos2x = cos * 2x sin * 2x cot * 2x = CSC * 2x-1 derivative of Cotx = - CSC * 2x
∫ (cos2x / sin ^ 2x) DX = ∫ cos * 2x sin * 2x / sin ^ 2xdx is simplified to = ∫ cot * 2xdx - ∫ 1dx = ∫ CSC * 2xdx - ∫ 1dx - ∫ 1dx = - cotx-2x + C

∫cos2x/sin^2xcos^2x dx On the first floor How did you figure this out on the second floor?

The answer is - 2 / sin2x

Find the indefinite integral ∫ sin3x cos2x DX

∫ sin3x cos2x dx =-cos5x/10-cosx/2+c