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You would probably use a power-reduction trig identity to solve this equation.
This states that sin2(x) = (1 - cos(2x))/2
Therefore, sin2(2x) = (1 - cos(4x))/2, or (1/2)(1 - cos(4x))
So, ∫ (1/2)(1 - cos(4x)) dx = (1/2) ∫ (1 - cos(4x)) dx.
Then, ∫ (1-cos4x)dx = x - (1/4)sin(4x) + c
Now, multiply that by (1/2) to get:
(x - (sin(4x)/4) + c)/2
Since c is an arbitrary constant, we have:
½(x - sin(4x) / 4) + c
OR
1/8 * (4x - sin(4x)) + c
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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)
Integration of sin2x?
Integral( sin(2x)dx) = -(cos(2x)/2) + C
Integral of tan square x secant x?
convert tan^2x into sin^2x/cos^2x and secant x into 1/cos x combine terms for integral sin^2x/cos^3x dx then sub in u= cos^3x and du=-2sin^2x dx
Integral of sec2x-cosx plus x2dx?
I wasn't entirely sure what you meant, but if the problem was to find the integral of [sec(2x)-cos(x)+x^2]dx, then in order to get the answer you must follow a couple of steps:First you should separate the problem into three parts as you are allowed to with integration. So it becomes the integral of sec(2x) - the integral of cos(x) + the integral of x^2Then solve each part separatelyThe integral of sec(2x) is -(cos(2x)/2)The integral of cos(x) is sin(x)The integral of x^2 isLastly you must combine them together:-(cos(2x)/2) - sin(x) + (x^3)/3
Integral of sin x sin 2x?
Ok, I know that sin 2x can be substituted out for 2 sin x cos x. so now I have the Integral of (sin x ) ( 2 sin x cos x ) dx which is 2 sin2x cos x dx If I use integration by parts with the u and dv, I find myself right back again..can you help. The book gives an answer, but I am not sure how it was achieved. the book gives 2/3 sin 2x cos x - 1/3 cos 2x sin x + C I may be making this more difficult than it is ? when you get the integral of 2 sin2x cos x dx use u substitution. u= sinx du= cosxdx. Then you'll get the integral of 2u^2 du.. .and then integrate...
How would you solve the integral of 1 plus tan2x plus tan squared 2x?
Integral of 1 is x Integral of tan(2x) = Integral of [sin(2x)/cos(2x)] =-ln (cos(2x)) /2 Integral of tan^2 (2x) = Integral of sec^2(2x)-1 = tan(2x)/2 - x Combining all, Integral of 1 plus tan(2x) plus tan squared 2x is x-ln(cos(2x))/2 +tan(2x)/2 - x + C = -ln (cos(2x))/2 + tan(2x)/2 + C
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 indefinte integral of sin2x?
The indefinite integral of sin 2x is -cos 2x / 2 + C, where C is any constant.
Integration of sin2x?
Integral( sin(2x)dx) = -(cos(2x)/2) + C
What is the integral of 1 divided by 2x squared?
0.5
Integral of tan square x secant x?
convert tan^2x into sin^2x/cos^2x and secant x into 1/cos x combine terms for integral sin^2x/cos^3x dx then sub in u= cos^3x and du=-2sin^2x dx
What is (1 cos x)(1- cos x)?
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
Integral of sec2x-cosx plus x2dx?
I wasn't entirely sure what you meant, but if the problem was to find the integral of [sec(2x)-cos(x)+x^2]dx, then in order to get the answer you must follow a couple of steps:First you should separate the problem into three parts as you are allowed to with integration. So it becomes the integral of sec(2x) - the integral of cos(x) + the integral of x^2Then solve each part separatelyThe integral of sec(2x) is -(cos(2x)/2)The integral of cos(x) is sin(x)The integral of x^2 isLastly you must combine them together:-(cos(2x)/2) - sin(x) + (x^3)/3
Integral of sine squared x?
.5(x-sin(x)cos(x))+c
Integral of sin x sin 2x?
Ok, I know that sin 2x can be substituted out for 2 sin x cos x. so now I have the Integral of (sin x ) ( 2 sin x cos x ) dx which is 2 sin2x cos x dx If I use integration by parts with the u and dv, I find myself right back again..can you help. The book gives an answer, but I am not sure how it was achieved. the book gives 2/3 sin 2x cos x - 1/3 cos 2x sin x + C I may be making this more difficult than it is ? when you get the integral of 2 sin2x cos x dx use u substitution. u= sinx du= cosxdx. Then you'll get the integral of 2u^2 du.. .and then integrate...
What does the graph of sin squared x look like?
cos(2x) = 1 - 2(sin(x)^2), so sin(x)^2 = 1/2 - 1/2*cos(2x).
What does cos squared x - Sin squared x equal?
2 x cosine squared x -1 which also equals cos (2x)
