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The method to use is 'integration by parts';
set u =x; du=dx; dv = sin(pi x)dx; v = cos(pi x)/pi.
so integral(u dv) = u*v - integral(v du)
then repeat the process.
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What is the relation between definite integrals and areas?
Consider the integral of sin x over the interval from 0 to 2pi. In this interval the value of sin x rises from 0 to 1 then falls through 0 to -1 and then rises again to 0. In other words the part of the sin x function between 0 and pi is 'above' the axis and the part between pi and 2pi is 'below' the axis. The value of this integral is zero because although the areas enclosed by the parts of the function between 0 and pi and pi and 2pi are the same the integral of the latter part is negative. The point I am trying to make is that a definite integral gives the area between a function and the horizontal axis but areas below the axis are negative. The integral of sin x over the interval from 0 to pi is 2. The integral of six x over the interval from pi to 2pi is -2.
Integral of x cosx dx?
The integral of x cos(x) dx is cos(x) + x sin(x) + C
What is the derivative of cos pi x?
-(pi)*sin(pi*x)
What is the period of y equals sin5x?
The period of y=sin(x) is 2*pi, so sin(x) repeats every 2*pi units. sin(5x) repeats every 2*pi/5 units. In general, the period of y=sin(n*x) is 2*pi/n.
How do you integrate periodic functions?
Same as any other function - but in the case of a definite integral, you can take advantage of the periodicity. For example, assuming that a certain function has a period of pi, and the value of the definite integral from zero to pi is 2, then the integral from zero to 2 x pi is 4.
How determine the area of curve under the x-axis?
Integrate the function for the curve, as normal, but the change the sign of the result. Be very careful that the curve is always on the same side of the x-axis between the limits of integration. If necessary, partition the integral. For example, to find the area between the x-axis and sin(x) between x=0 and x=3*pi, you will need Integral of sin(x) between 0 and pi, -[integral of sin(x) between pi and 2*pi] - this is where the curve is below the x-axis. +integral of sin(x) between 2*pi and 3*pi.
What is the relation between definite integrals and areas?
Consider the integral of sin x over the interval from 0 to 2pi. In this interval the value of sin x rises from 0 to 1 then falls through 0 to -1 and then rises again to 0. In other words the part of the sin x function between 0 and pi is 'above' the axis and the part between pi and 2pi is 'below' the axis. The value of this integral is zero because although the areas enclosed by the parts of the function between 0 and pi and pi and 2pi are the same the integral of the latter part is negative. The point I am trying to make is that a definite integral gives the area between a function and the horizontal axis but areas below the axis are negative. The integral of sin x over the interval from 0 to pi is 2. The integral of six x over the interval from pi to 2pi is -2.
Integral of 1 divided by sinx cosx?
Integral of [1/(sin x cos x) dx] (substitute sin2 x + cos2 x for 1)= Integral of [(sin2 x + cos2 x)/(sin x cos x) dx]= Integral of [sin2 x/(sin x cos x) dx] + Integral of [cos2 x/(sin x cos x) dx]= Integral of (sin x/cos x dx) + Integral of (cos x/sin x dx)= Integral of tan x dx + Integral of cot x dx= ln |sec x| + ln |sin x| + C
Why sin integral is cos?
sin integral is -cos This is so because the derivative of cos x = -sin x
What is the derivative of cos pi x plus sin pi y all to the 8th power equals 44?
(cos(pi x) + sin(pi y) )^8 = 44 differentiate both sides with respect to x 8 ( cos(pi x) + sin (pi y ) )^7 d/dx ( cos(pi x) + sin (pi y) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (-sin (pi x) pi + cos (pi y) pi dy/dx ) = 0 8 ( cos(pi x) + sin (pi y ) )^7 (pi cos(pi y) dy/dx - pi sin (pi x) ) = 0 cos(pi y) dy/dx - pi sin(pi x) = 0 cos(pi y) dy/dx = sin(pi x) dy/dx = sin (pi x) / cos(pi y)
Integral of 9 pi?
This depends on what you are integrating with respect to. Let's assume: x. Integral of 9*pi = 9*pi*x + C. However, if you are integrating with respect to pi, then integral of 9*pi is (9/2)pi^2 + C
What is the derivative of sin pi x?
pi cos(pi x)
How do you find the volume of the sphere using integration?
Required:-Knowlege of how Integration works.-Knowlege of derivatives-Knowlege of Triganometry.-Knowlege of Basic Geometry.Given: The volume of a sqhere and be represented in infinitely small cross sections [dx].Each cross section looks like a circle [pi(r)2]If you made a graph of the radius of the cross sections from one side of the sphere to the other it would look like a half cirlce [Sin(u/a) from 0 to Pi where a is the inverse of the radius of the sqhere] (radius of the sphere=1=u/a=x=hypotenuse, radius of the crossection equals=opposite=sin(x), because sin(x)=opp/hyp=opp/1=opp. the center of the sphere is located at x=pi/2)Combine the above equations and integrate to find the sum of the cross sections.[integral from 0 to pi of (pi(sin(x))2dx)]=[integral from 0 to pi of (pi (1-cos(2x))/2 dx)]=[integral of (pi/2 dx)]-[integral of (pi/2)*(cos(2x))dx)]= (pi/2(x)-pi/4(sin(2x)) evlauated from 0 to pi.volume of a sphere V=4/3pi(r)3=[(pi/2)(u/a)-(pi/4)(sin2(u/a)) evaluated from 0 to (a)Pi
What is the integral of sin x?
-cos x + Constant
What is the definite integral from 0 to pi divided by 6 of sin cubed 2x cos 2x dx?
Let y = sin 2x Then dy/dx = 2*cos 2x Also, when x = 0, y = 0 and when x = pi/6, y = sin(2*pi/6) = sin(pi/3) = sqrt(3)/2 Therefore, the integral becomes definite integral from 0 to √3/2 of 1/2*y3dy = difference between 1/2*(y4)/4 = (y4)/8 evaluated at √3/2 and 0. = (√3/2)4 /8 = 9/128 = 0.0703 approx.
Integral of x cosx dx?
The integral of x cos(x) dx is cos(x) + x sin(x) + C
Sin x - cos x 0 0?
sin x - cos x = 0sin x = cos x(sin x)^2 = (cos x)^2(sin x)^2 = 1 - (sin x)^22(sin x)^2 = 1(sin x)^2 = 1/2sin x = ± √(1/2)sin x = ± (1/√2)sin x = ± (1/√2)(√2/√2)sin x = ± √2/2x = ± pi/4 (± 45 degrees)Any multiple of 2pi can be added to these values and sine (also cosine) is still ± √2/2. Thus all solutions of sin x - cos x = 0 or sin x = cos x are given byx = ± pi/4 ± 2npi, where n is any integer.By choosing any two integers , such as n = 0, n = 1, n = 2 we can find some solutions of sin x - cos x = 0.n = 0, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(0)(pi) = ± pi/4 ± 0 = ± pi/4n = 1, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(1)(pi) = ± pi/4 ± 2pi = ± 9pi/4n = 2, x = ± pi/4 ± (2)(n)(pi) = ± pi/4 ± (2)(2)(pi) = ± pi/4 ± 4pi = ± 17pi/4
