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Integral of sin squared x?
First we look at the double-angle identity of cos2x. We know that: cos2x = cos^2x - sin^2x cos2x = [1-sin^2x] - sin^2x.............. (From sin^2x + cos^2x = 1, cos^2x = 1 - sin^2x) Therefore: cos2x = 1 - 2sin^2x 2sin^2x = 1 - cos2x sin^2x = 1/2(1-cos2x) sin^2x = 1/2 - cos2x/2 And intergrating, we get: x/2 - sin2x/4 + c...................(Integral of cos2x = 1/2sin2x; and c is a constant)
What is the antiderivative of sine squared?
∫sin2x dxUse the identity sin2x = ½ - ½(cos2x)∫[½ - ½(cos2x)] dx = ∫½ dx - ∫½(cos2x) dxLet's split it up into ∫½ dx and ∫½(cos2x) dx∫½ dx = x/2 (we'll put the constant in at the end)∫½(cos2x) dx (Use u substitution with u=2x and du = 2 dx)∫cosu ¼du = ¼∫cosu du = ¼sinu + c = ¼sin2x (remember to resubstitute)Subtract the two parts and add a constantx/2 - ¼(sin2x) + cThis is also equivalent to: ½(x - sinxcosx) + c
What is cos2x equal to?
You can look up "trigonometric identities" in Wikipedia.Cos(2x), among other things, is equal to (cos x)^2 - (sin x)^2 If you meant cos squared x, or (cos x)^2, that is equal to (1 + cos(2x))/2
Does cos2x equal 2cosxsinx?
No; sin2x = 2 cosx sinx
What is the integral of xsin2xdx?
∫ xsin(2x) dx = (-1/2)xcos2x + (1/4)sin2x You get this by using Integration by Parts. An integral in the form ∫udv can be written as uv-∫vdu In the case of your problem u=x, du=1, dv=sin2x, v=(-1/2)cos2x <--You get v by integrating dv Using the formula ∫udv = uv- ∫vdu and by plugging in what has been defined above you get ∫xsin(2x)dx = (-1/2)xcos2x - ∫(-1/2)cos2x(1) By integrating ∫(-1/2)cos2x, you get (-1/4)sin2x. When you plug that back in, you get ∫xsin2xdx=(-1/2)xcos2x-(-1/4)sin2x or just simply ∫xsin(2x)dx = (-1/2)xcos(2x) + (1/4)sin(2x)
How do you verify cos2 theta divided by csc2 theta plus cos4 theta equals cos2 theta?
Using x instead of theta, cos2x/cosec2x + cos4x = cos2x*sin2x + cos4x = cos2x*(sin2x + cos2x) = cos2x*1 = cos2x
How do you prove one - tan square x divided by one plus tan square xequal to cos two x?
(1 - tan2x)/(1 + tan2x) = (1 - sin2x/cos2x)/(1 + sin2x/cos2x) = (cos2x - sin2x)/(cos2x + sin2x) = (cos2x - sin2x)/1 = (cos2x - sin2x) = cos(2x)
How do you factor sinx-cos2x-1?
[sinx - cos2x - 1] is already factored the most it can be
Integral of sin squared x?
First we look at the double-angle identity of cos2x. We know that: cos2x = cos^2x - sin^2x cos2x = [1-sin^2x] - sin^2x.............. (From sin^2x + cos^2x = 1, cos^2x = 1 - sin^2x) Therefore: cos2x = 1 - 2sin^2x 2sin^2x = 1 - cos2x sin^2x = 1/2(1-cos2x) sin^2x = 1/2 - cos2x/2 And intergrating, we get: x/2 - sin2x/4 + c...................(Integral of cos2x = 1/2sin2x; and c is a constant)
How do you solve 2 cos squared x - sinx - 1?
1
Can you prove that cossquaredx - sinsquaredx equals 2cossquaredx -1?
3
What is the antiderivative of sine squared?
∫sin2x dxUse the identity sin2x = ½ - ½(cos2x)∫[½ - ½(cos2x)] dx = ∫½ dx - ∫½(cos2x) dxLet's split it up into ∫½ dx and ∫½(cos2x) dx∫½ dx = x/2 (we'll put the constant in at the end)∫½(cos2x) dx (Use u substitution with u=2x and du = 2 dx)∫cosu ¼du = ¼∫cosu du = ¼sinu + c = ¼sin2x (remember to resubstitute)Subtract the two parts and add a constantx/2 - ¼(sin2x) + cThis is also equivalent to: ½(x - sinxcosx) + c
How do you simplify sec x tan x parenthesees one minus sin squared x?
You use the identity sin2x + cos2x = 1 (to simplify the expression in parentheses), and convert all functions to sines and cosines. sec x tan x (1 - sin2x) = (1/cos x) (sin x / cos x) (cos2x) = (sin x / cos2x) cos2x = sin x
Derivative of tanx?
It is sec2x, this is the same as 1/cos2x.
How do you solve sin squared x divided by 1 - cos x?
Use this identity sin2x+cos2x=1 sin2x=1-cos2x so sin2x/(1-cosx) =(1-cos2x)/(1-cosx) =(1-cosx)(1+cosx)/(1-cosx) =1+cosx
How does sin2x divided by 1-cosx equal 1 plus cosx?
sin2x / (1-cos x) = (1-cos2x) / (1-cos x) = (1-cos x)(1+cos x) / (1-cos x) = (1+cos x) sin2x=1-cos2x as sin2x+cos2x=1 1-cos2x = (1-cos x)(1+cos x) as a2-b2=(a-b)(a+b)
What is cos2x equal to?
You can look up "trigonometric identities" in Wikipedia.Cos(2x), among other things, is equal to (cos x)^2 - (sin x)^2 If you meant cos squared x, or (cos x)^2, that is equal to (1 + cos(2x))/2
