Indefinite integral of cos2x / SiNx DX

Indefinite integral of cos2x / SiNx DX

The solution is as follows:
∫cos2x/sinx dx
=∫[1-2(sinx)^2]/sinx dx
=∫cscxdx-∫2sinxdx
=∫cscx(cscx-ctgx)/(cscx-ctgx)dx+2cox
=∫1/(cscx-ctgx)d(cscx-ctgx)+2cosx
=ln(cscx-ctgx)+2cosx+C
The above answers are for reference only,

Radical 3 / 2sin2x-1 + cos2x / 2-1 / 2 How to solve the unity formula?

(√3/2)sin2x-1+(1/2)cos2x-1/2
=cos(30)sin2x+sin30cos2x-3/2
=sin(2x+30)-3/2

The minimum positive period of y = 1 / 2 sin2x - the root of 2 / 2 cos2x is

According to the meaning of the title: y = 1 / 2sin2x - √ 3 / 2cos2x = sin2xcos π / 3-cos2xsin π / 3 = sin (2x - π / 3) ν the minimum positive period T = 2 π / 2 = π or by conventional method: asin α + bcos α = √ a + bsin (ax + β), where β = arctan (B / a)

Reduction of radical 3 / 2sin2x-1 / 2cos2x + 1 / 2

=cospi/6sin2x-sinpi/6cos2x+1/2
=sin(2x-pi/6)+1/2

It is known that cos2x = 4 / 5, and X belongs to (7 π, 2 π) of (1) sin ^ 4x + cos ^ 4x (2) tan Wrong number. The second question is tan2 X

cos2x=cos²x-sin²x=4/5
7π/2

If TaNx = 2, sinxcosx + cos2x =?

sinx/cosx=tanx=2
sinx=2cosx
The identity sin? X + cos? X = 1 is introduced
cos²x=1/5
sinxcosx=(2cosx)cosx=2cos²x=2/5
cos2x=2cos²x-1=-3/5
So the original formula = - 1 / 5

Given TaNx = 2, then the value of 2sin2x-sinxcosx + cos2x is_____ Teacher, please come in, I want the correct answer, incorrect please go back

three-fifths

Given TaNx = 2, find 2sin2x-sinxcosx cos2x

It is divided into two parts
Note: sin2x = 2sinxcosx
=4sinxcosx / {(cosx) ^ 2 + (SiNx) ^ 2} note: {(cosx) ^ 2 + (SiNx) ^ 2 = 1
=4 TaNx / {1 + (TaNx) ^ 2} note: the numerator and denominator are divided by (cosx) ^ 2 at the same time
Sinxcosx cos2x = 1 / 2sin2xcos2x note: sinxcosx = 1 / 2sin2x
=Note: {(cos2x) ^ 2 + (sin2x) ^ 2 = 1
=1 / 2tan2x / {1 + (TaNx) ^ 2} note: the numerator and denominator are divided by (cos2x) ^ 2 at the same time
Tan = 2 tan2x = (2tanx) / {1 - (TaNx) ^ 2} = (4 / 3) the number can be substituted into the calculation
2sin2x-sinxcosx cos2x=26/15

In the case of TaNx = 3 (sinxcosx + cos2x) =?

tanx=3 cosx^2=1/(1+tanx^2)=1/10
sinxcosx+cos2x=tanx*cosx^2+2cosx^2-1=3/10+2/10-1=-1/2

1-cos2x =? Formula

cos2x=cos²x-sin²x=2cos²x-1=1-2sin²x
You can choose the formula you need according to the situation
So 1-cos2x = 1 - (1-2sin? X) = 2Sin? X