Using cos α Represents sin α (quartic) - Sin α (quadratic) + cos α (quadratic)

Using cos α Represents sin α (quartic) - Sin α (quadratic) + cos α (quadratic)

∵sin ²α= 1-cos ²α ∴sin⁴ α- sin ²α+ cos ²α= (1-coos ²α)²- (1-cos ²α)+ cos ²α= 1-2cos ²α+ cos⁴ α- 1+2cos ²α= cos⁴ α

Proof: cos quartic X / 2 - Sin quartic X / 2 = cosx

Following formula: cos square x-sin square x = cos2x; Sum formula: cos square x + sin square x = 1. Then the left formula = (COS square X / 2-sin square X / 2) * (COS square X / 2 + sin square X / 2) = cos square X / 2 - Sin square X / 2 = cosx

How to calculate ∫ x ^ 2 √ 1 + x ^ 2DX integral?

Let x = Tan θ, dx = sec ²θ d θ ∫ x ² √(1 + x ²) dx= ∫ tan ²θ * | sec θ| * (sec ²θ d θ)= ∫ tan ²θ sec ³θ d θ= ∫ (sec ²θ - 1)sec ³θ d θ= ∫ sec⁵ θ d θ - ∫...

9.6 finding indefinite integral ∫ [cos (x / 2)] ^ 2DX

∫ [cos(x/2)]^2dx
=1/2*∫[2(cosx/2)^2-1+1 ]dx
=1/2*∫cosxdx+1/2*∫dx
=1/2*sinx+x/2+C

How to solve ∫ cos (x-1) DX, ∫ x ^ 3E ^ x ^ 2DX

∫ cos(x - 1) dx= ∫ cos(x - 1) d(x - 1)= sin(x - 1) + C∫ x ³ e^(x ²) DX -- > make u = x ², du = 2x dx-->= ∫ uxe^u · du/(2x)= (1/2)∫ ue^u du = (1/2)∫ u de^u= (1/2)ue^u - (1/2)∫ e^u du= (1/2)...

∫ x ^ 2DX ^ 2, how?

Make u = x ^ 2
Then ∫ x ^ 2DX ^ 2 = ∫ UDU = u ^ 2 / 2 = x ^ 4 / 2