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There is some kind of formula here, half angle, or some such that I forget, but I do remember the algorithm. So...,
int[cos(10X)cos(15X)] dx
since this is multiplicative, switch it around
int[cos(15X)cos(10X)] dx
int[cos(15X - 10X)/2(15 -10) + cos(15X + 10X)/2(15 + 10)] dx
int[cos(5X)/10 + cos(25X)/50] dx
= 1/10sin(5X) + 1/50sin(25X) + C
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What else can I help you with?
Integrating Cos4 x dx?
To integrate such a function, you must use the u-substitution technique. Let u = 4x; therefore du = 4 dx. Then du/4 = dx. The integral becomes : Int(cos 4x dx) = Int[(cos u)(du/4)] = ¼ Int(cos u du) = ¼ (sin u) + C = ¼ sin 4x + C
What is the integral of sin x cubed?
= cos(x)-(cos3(x))/3 * * * * * Right numbers, wrong sign! Int(sin3x)dx = Int(sin2x*sinx)dx = Int[(1-cos2x)*sinx]dx = Int(sinx)dx + Int[-cos2x*sinx]dx Int(sinx)dx = -cosx . . . . . (I) Int[-cos2x*sinx]dx Let u = cosx, the du = -sinxdx so Int(u2)du = u3/3 = 1/3*cos3x . . . . (II) So Int(sin3x)dx = 1/3*cos3x - cosx + C Alternatively, using the multiple angle identities, you can show that sin3x = 1/4*[3sinx - sin3x] which gives Int(sin3x)dx = 1/4*{1/3*cos(3x) - 3cosx} + C
Integration of tangent x?
To integrate tan(x), you must break up tangent into sine over cosine, with that being done, all you have is a u-substitution with the cosine. This should give: int(tan(x)dx)=int(sin(x)/cos(x)dx)=int(-(1/u)*du)=-ln|u|+C=-ln|cos(x)|+C u=cos(x) du=-sin(x)dx
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
What is the integral of sin squared x times cos squared x?
∫sin²x cos²x dx = ∫(1-cos²x)cos²x dx =∫cos²xdx-⌠cos²xcos²xdx =1/2⌠1+cos2x dx-1/2⌠[(1+cos2x)(1+cos2x)] Do the operations, distributions, arrange common numbers, and try to sort out the factors as a polynomiom. Then, =1/2x+1/4sin2x-1/2x-1/2sin2x-1/4x-1/16sin4x =-1/4x-1/16sin4x
What is the integral of sin 5X cos 10X dX?
use product-to-sumformula sin u cos v =1/2 [sin(u+v)+ sin(u-v)]so you get1/2 int: (sin 15x) - (sin5x) dxsplit it1/2 int: sin 15x dx- 1/2 int: sin 5x dxusing substitution you can conclude that1/30 int: sin u du- 1/10 in sin w dw(you get the fraction change when you set dx=duand dw)so then- (cos u)/30 + (cos w)/10replace the substitution(cos 5x)/10 - (cos 15x)/30 + Constant
Integrating Cos4 x dx?
To integrate such a function, you must use the u-substitution technique. Let u = 4x; therefore du = 4 dx. Then du/4 = dx. The integral becomes : Int(cos 4x dx) = Int[(cos u)(du/4)] = ¼ Int(cos u du) = ¼ (sin u) + C = ¼ sin 4x + C
Can you integrate -cos x plus c?
Why not? Just a second integration. Drop the constant. int[- cos(x)] dx the negative implies - 1 and can be brought out side the integrand - int[cos(x)] dx = - sin(x) + C ==========
Integral of (sin2x)(cos3x)?
Int[sin(2x)*cos(3x)]dx = Int[(2sinx*cosx)*(4cos^3x - 3cosx)]dx= Int[(8sinx*cos^4x - 6sinx*cos^2x)]dx Let cosx = u then du/dx = -sinx So, the integral is Int[-8*u^4 + 6*u^2]du = -8/5*u^5 + 2u^3 + c where c is a constant of integration = -8/5*cos^5x + 2cos^3x + c
Int sin 3x cos 5x dx?
integral sin(3 x) cos(5 x) dx = 1/16 (8 cos^2(x)-cos(8 x))+C
What is the integral of sin x cubed?
= cos(x)-(cos3(x))/3 * * * * * Right numbers, wrong sign! Int(sin3x)dx = Int(sin2x*sinx)dx = Int[(1-cos2x)*sinx]dx = Int(sinx)dx + Int[-cos2x*sinx]dx Int(sinx)dx = -cosx . . . . . (I) Int[-cos2x*sinx]dx Let u = cosx, the du = -sinxdx so Int(u2)du = u3/3 = 1/3*cos3x . . . . (II) So Int(sin3x)dx = 1/3*cos3x - cosx + C Alternatively, using the multiple angle identities, you can show that sin3x = 1/4*[3sinx - sin3x] which gives Int(sin3x)dx = 1/4*{1/3*cos(3x) - 3cosx} + C
Integration of xtanx?
To integrate ( x \tan(x) ), we can use integration by parts. Let ( u = x ) and ( dv = \tan(x) , dx ). This gives ( du = dx ) and ( v = -\ln|\cos(x)| ). Applying the integration by parts formula ( \int u , dv = uv - \int v , du ), we obtain: [ \int x \tan(x) , dx = -x \ln|\cos(x)| + \int \ln|\cos(x)| , dx + C ] The integral ( \int \ln|\cos(x)| , dx ) does not have a simple closed form, so the final result may be expressed in terms of this integral along with the logarithmic term.
Integration of cos power of 2x plus sin2x dx?
To integrate the expression ( \cos^2(2x) + \sin(2x) , dx ), we can first rewrite ( \cos^2(2x) ) using the Pythagorean identity: ( \cos^2(2x) = \frac{1 + \cos(4x)}{2} ). The integral then becomes: [ \int \left( \frac{1 + \cos(4x)}{2} + \sin(2x) \right) , dx. ] This simplifies to: [ \frac{1}{2} \int (1 + \cos(4x)) , dx + \int \sin(2x) , dx, ] which can be integrated to yield: [ \frac{1}{2} \left( x + \frac{\sin(4x)}{4} \right) - \frac{1}{2} \cos(2x) + C, ] where ( C ) is the constant of integration.
What is the integral of cosxsinx?
The integral of ( \cos x \sin x ) can be computed using a trigonometric identity. We use the identity ( \sin(2x) = 2 \sin x \cos x ), which allows us to rewrite the integral as: [ \int \cos x \sin x , dx = \frac{1}{2} \int \sin(2x) , dx. ] Integrating ( \sin(2x) ) gives: [ \frac{-1}{2} \cos(2x) + C, ] thus the final result is: [ \int \cos x \sin x , dx = \frac{-1}{4} \cos(2x) + C. ]
How would you evaluate the indefinite integral -2xcos3xdx?
Integrate by parts: ∫ uv dx = u ∫ v dx - ∫ (u' ∫ v dx) dx Let u = -2x Let v = cos 3x → u' = d/dx -2x = -2 → ∫ -2x cos 3x dx = -2x ∫ cos 3x dx - ∫ (-2 ∫ cos 3x dx) dx = -2x/3 sin 3x - ∫ -2/3 sin 3x dx = -2x/3 sin 3x - 2/9 cos 3x + c
Integration of tangent x?
To integrate tan(x), you must break up tangent into sine over cosine, with that being done, all you have is a u-substitution with the cosine. This should give: int(tan(x)dx)=int(sin(x)/cos(x)dx)=int(-(1/u)*du)=-ln|u|+C=-ln|cos(x)|+C u=cos(x) du=-sin(x)dx
Double integral of x siny dx dy?
We have:int int (x * sin(y)) dx dyIntegrate x first:int(x)dx = 1/2 * x2 + CNow integrate sin(y):int(sin(y))dy = -cos(y) + CMultiply:-1/2 * x2 * cos(y) + C
