Solving indefinite integrals 1. ʃ x ^ 4 / (x? + 1) DX 2. ʃ (1 + cos? X) / (1 + cos2x) DX

Solving indefinite integrals 1. ʃ x ^ 4 / (x? + 1) DX 2. ʃ (1 + cos? X) / (1 + cos2x) DX

1、 =∫(x^4+x²-x²-1+1)/(x²+1)dx=∫[x²-1+1/(x²+1)]dx=(1/3)x³-x+arctanx+C
2、 =∫(1+cos²x)/(2cos²x)dx=(1/2)∫[1+1/cos²x]dx=(1/2)x+(1/2)tanx+C

∫cos(1-cos2x)dx=

∫cos(1-cos2x)dx
=∫2sin^xcosxdx
=∫2sin^xdsinx
=2/3 sin³x+C

What is the maximum value of y = cos ^ 2x + cos2x?

y=(cosx)^2+cos2x
=(cosx)^2+2(cosx)^2-1=3(cosx)^2-1
When (cosx) ^ 2 = 1, cosx = ± 1, x = k Π
Y Max is 3-1 = 2

Y = the maximum value of cos2x + cos (2x + π / 3)

The third π ≈ 3.14 × 1 / 3 ≈ 1.05
Y=COS2X+COS(2X+π／3)
　=COS2X·（2X+π／3)
　=COS(4X^2+X2π／3)
　=COS4·〔（X＋π／12)^2－π^2／144〕
Then draw the highest point of the graph of the equation, x = - π / 12, and substitute it into the equation to get the maximum value of - cos π ^ 2 / 38
(to be clear, I haven't touched a math book for a long time after work. I made it by memory. If you think it's right, you can adopt it. If it's not right, I'm sure it's right. The general steps are as follows.)

If y = [(1 + TaNx) / cos 2 x] / (cos2x + sin2x) is defined as (0, π / 4), what is the range of the function

f(x)=(cos²x+sinxcosx)/(sin2x+cos2x)
=(1/2+1/2cos2x+1/2sin2x)/(sin2x+cos2x)
=(1/2)+(1/2)/(sin2x+cos2x)
=(1/2)+1/[2√2sin(2x+π/4)]
Zero

If TaNx = cos (π / 2 + x), then the value of sin2x + cos2x is given

tanx=cos(π/2+x)=-sinx
sinx/cosx=-sinx
cosx=-1
sinx=0
sin2x+cos2x=2sinxcosx+cos²x-sin²x=1

F (x) = radical 3sin2x-2sin ^ 2x (1) If the point P (1, - radical 3) is on the middle edge of the angle α, find the value of F (α) (2) If x ∈ [- π / 6], find the range of F (x) The second problem is to find the range of F (x) if x ∈ [- π / 6, π / 3]

(1) If a = - 60, sin (a) = - radical 3 / 2, sin (2a) = - radical 3 / 2, then f (a) = - 3 (2) f (x) = root 3sin (2x) + cos (2x) - 1 = 2Sin (2x + π / 6) - 1x ∈ [- π / 6, π / 3], then 2x + π / 6 ∈ [- π / 6, π 5 / 6], sin (2x + π / 6) ∈ [- 1 / 2,1] f (x) ∈ [- 2,1]

F (x) = 2Sin ^ 2x + radical 3sin2x + α, and | f (x)|

F (x) = 2Sin ^ 2x + Radix 3sin2x + α = (2 + radical 3) sin2x + α
|f(x)|

Given the function f (x) = 2Sin ^ 2x + root sign 3sin2x, find the minimum positive period of function f (x). (2) f (x) is less than m + 2 when x belongs to (2) If f (x) is less than m + 2, it is always true that x belongs to [0, π / 2]. We can find the value range of real number M

F (x) = 2Sin ^ 2x + radical 3sin2x
=1-cos2x+√3sin2x
=1+2（-1/2cos2x+√3/2 sin2x）
=1+2（-sin30°cos2x+cos30°sin2x）
=1+2sin（2x-30°）
So the minimum positive period T = 2 π / 2 = π
m> 2 sin (2x-30 °) - 1, 2Sin (2x-30 °) - 11, Heng holds

F (x) = negative radical 3sin2x-2sin square x If the point P (1, - radical 3) is on the final edge of the angle a, find f (a)

From the meaning of the title
Tana = - √ 3, a = k π - π / 3
f(a)=-√3sin2（-π/3)+cos2(-π/3)-1
=3/2-1/2-1
=0