Given that f (SiNx) = 1-cos2x, how much is f (cosx) equal,

Given that f (SiNx) = 1-cos2x, how much is f (cosx) equal,

f（sinx）=1-cos2x=2(sin x)^2,
The existing f (x) = 2x ^ 2,
So f (COS x) = 2 * (COS x) ^ 2 = 1 + cos2x

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

Indefinite integral: ∫ (cos2x) / (SiNx) ^ 2. Cosx ^ 2

∫(cos2x)/(sinX)^2.cosx^2 dx
＝∫（cosx^2-sinx^2)/(sinX)^2.cosx^2 dx
＝∫（1/sinx^2-1/cosx^2) dx
=∫(cscx^2-secx^2) dx
=-cotx-tanx+C

What is the indefinite integral of cos2x divided by (cosx squared * SiNx squared)

∫ cos2x / (cos²x * sin²x) dx
= ∫ cos2x / (cosx*sinx)² dx
= ∫ cos2x / (1/2 * 2sinx*cosx)² dx
= ∫ cos2x / [(1/2)² * (sin2x)²] dx
= ∫ cos2x / [(1/4) * sin²2x] dx
= 4∫ cos2x / sin²2x dx
= (4/2)∫ cos2x / sin²2x d(2x)
= 2∫ d(sin2x) / (sin2x)²
= -2/sin2x + C
= -2csc2x + C

Find the indefinite integral of cos2x / (cosx * SiNx) ^ 2

∫cos2x / (cosx*sinx)^2 = 4∫ cos2x/sin²2x dx
=4∫ cot2x * csc2x dx
=-2∫ d csc2x
=-2csc2x +C

Find the indefinite integral ∫ [(cosx) ^ 4 / (SiNx) ^ 3] DX

The first equal sign and the fourth equal sign are wrong. You can do it step by step. The mobile phone is not easy to call. Let me tell you the steps (using ζ to indicate the integral number below): write (cosx) ^ 4 into (1-sin ^ x) ^, the original formula = ζ sinxdx-2 ζ (1 / SiNx) DX + ζ [1 / (SiNx) ^ 3] DX

Intermediate steps of ∫ (cos2x - (COS ^ 3) 2x) DX = ∫ (sin ^ 2) 2x * 1 / 2D (sin2x) What's the middle step? How can two subtractive integrals become multiplication

∫ cos2x(1-cos²2x)dx
=∫ sin²2x*cos2xdx
=∫ sin²2x*½d(sin2x)

Find the indefinite integral (1 / sin ^ 2xcos ^ 2x) DX

The original formula = ∫ 4DX / (2sinxcosx) 2
=4∫dx/sin²2x
=2∫csc²2xd2x
=-2cot2x+C

Why sin (2x + π / 2) = cos2x

Expansion sin (2x + π / 2) = sin (2x) cos (π / 2) + cos (2x) sin (π / 2)
=cos2x
Get the certificate

How does cos2x become sin (2x + π / 2)?

Induction formula sin (a + π / 2) = cosa
So sin (2x + π / 2) = cos2x