How to calculate the integral of 1 / ((SiNx + cosx) ^ 2)? RT

How to calculate the integral of 1 / ((SiNx + cosx) ^ 2)? RT

The original formula = ∫ 1 / [√ 2Sin (x + π / 4)] & sup2; DX
=1/2∫dx/sin²(x+π/4)
=1/2∫csc²(x+π/4)d(x+π/4)
=-1/2∫-csc²(x+π/4)d(x+π/4)
=-1/2*cot(x+π/4)+C

What is the integral of (SiNx * cosx) ^ 2?

∫ (SiNx * cosx) & sup2; DX = 1 / 4 ∫ Sin & sup2; 2x DX uses sin2x = 2sinx * cosx = 1 / 8 ∫ (1 - cos4x) DX uses 2Sin & sup2; X = 1 - cos2x = 1 / 8 [x - sin4x / 4] + C = x / 8 - sin4x / 32 + C

Finding the maximum and minimum of y = 1-2cosx + 1 / 2 * (SiNx) ^ 2

Y = 1-2cosx + 1 / 2 * (SiNx) ^ 2 = 1-2cosx + 1 / 2 * (1-cos & # 178; x) = - 1 / 2cos & # 178; x-2cosx + 3 / 2 = - 1 / 2 (COS & # 178; X + 4cosx) + 3 / 2 = - 1 / 2 (cosx + 2) &# 178; + 7 / 2 when cosx = 1, the function has the minimum value = - 9 / 2 + 7 / 2 = - 1; when cosx = - 1, the function has the maximum value = - 1 / 2 + 7 / 2 = 3

Y = (SiNx + cosx) ^ 2 + 2cosx ^ 2

y=sin²x+cos²x+2sinxcosx+2(1+cos2x)/2
=sin2x+cos2x+2
=√2sin(2x+π/4)+2
So the maximum value = √ 2 + 2
Minimum = √ 2 / 2
The decreasing order is 2K π + π / 2