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Integration of cos2x?
Using chain rule:integral of cos2x dx= 1/2 * sin2x + C
Why does -2sinxcosx equal -sin2x?
-4
Sin2x - radical 2 cosx equals 0?
Sin2x = radical 2
What is the integration of cotx?
The integral of cot (x) dx is ln (absolute value (sin (x))) + C. Without using the absolute value, you can use the square root of the square, i.e. ln (square root (sin2x)) + C
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)
Integration of cos2x?
Using chain rule:integral of cos2x dx= 1/2 * sin2x + C
Is sin2x equal to sin2x?
sin2X = sin2X What is it about ' equation ' you do you not understand. Of course they are equal!
Why does -2sinxcosx equal -sin2x?
-4
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)
Sin2x - radical 2 cosx equals 0?
Sin2x = radical 2
What value is equivalent to 1 - cosx squared?
sin2x because sin2x + cos2x = 1
What does sin2x - 1 equal?
1
How does 1 plus cot squared x equals csc squared x?
The proof of this trig identity relies on the pythagorean trig identity, the most famous trig identity of all time: sin2x + cos2x = 1, or 1 - cos2x = sin2x. 1 + cot2x = csc2x 1 = csc2x - cot2x 1 = 1/sin2x - cos2x/sin2x 1 = (1 - cos2x)/sin2x ...using the pythagorean trig identity... 1 = sin2x/sin2x 1 = 1 So this is less of a proof and more of a verification.
How do you find sin theta when tan theta equals one fourth?
I will note x instead of theta tan(x) = sin(x) / cos(x) = 1/4 sin(x) = cos(x)/4 = ±sqrt(1-sin2x)/4 as cos2x + sin2 x = 1 4 sin(x) = ±sqrt(1-sin2x) 16 sin2x = 1-sin2x 17 sin2x = 1 sin2x = 1/17 sin(x) = ±1/sqrt(17)
What is the integration of cotx?
The integral of cot (x) dx is ln (absolute value (sin (x))) + C. Without using the absolute value, you can use the square root of the square, i.e. ln (square root (sin2x)) + C
Which is equivalent to (1- sin2x)tanx?
-1
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)
