0
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
=========================
Still curious? Ask our experts.
Chat with our AI personalities
Ezra
Professor
TaigaAdd your answer:
Earn +20 pts
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
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
Integral of cosine squared x?
Integral of cos^2x=(1/2)(cosxsinx+x)+CHere is why:Here is one method: use integration by parts and let u=cosx and dv=cosxdxdu=-sinx v=sinxInt(udv)=uv-Int(vdu) so uv=cosx(sinx) and vdu=sinx(-sinx)so we have:Int(cos^2(x)=(cosx)(sinx)+Int(sin^2x)(the (-) became + because of the -sinx, so we add Int(vdu))Now it looks not better because we have sin^2x instead of cos^2x,but sin^2x=1-cos^2x since sin^2x+cos^2x=1So we haveInt(cos^2x)=cosxsinx+Int(1-cos^2x)=cosxsinx+Int(1)-Int(cos^2x)So now add the -Int(cos^2x) on the RHS to the one on the LHS2Int(cos^2x)=cosxsinx+xso Int(cos^2x)=1/2[cosxsinsx+x] and now add the constant!final answerIntegral of cos^2x=(1/2)(cosx sinx + x)+C = x/2 + (1/4)sin 2x + C(because sin x cos x = (1/2)sin 2x)Another method is:Use the half-angle identity, (cos x)^2 = (1/2)(1 + cos 2x). So we have:Int[(cos x)^2 dx] = Int[(1/2)(1 + cos 2x)] dx = (1/2)[[Int(1 dx)] + [Int(cos 2x dx)]]= (1/2)[x + (1/2)sin 2x] + C= x/2 +(1/4)sin 2x + C
What is the integral of 10x?
∫ 10x dx Factor out the constant: 10 ∫ x dx Therefore, by the power rule, we obtain: 10x(1 + 1)/(1 + 1) + k = 10x²/2 + k = 5x² + k
Find the perimeter of the cord ra1 cos?
r=a(1+cos x) r^2=a^2(1+cos x)^2 = a^2 + 2[a^2][cos(x)] + [a^2][cos^2(x)] dr/dx = r' = -a sin(x) (r')^2 = [a^2][sin^2 (x)] Therefore perimeter (s) of curve r=a(1+cos x) in polar coordinate with x vary from 0 to Pi (due to curve is symmetry on axis x=0) is s = 2 Int { sqrt[r^2 + (r')^2] } dx where x vary from 0 to Pi. Thus sqrt[r^2 + (r')^2] = sqrt { a^2 + 2[a^2][cos(x)] + [a^2][cos^2 (x)] + [a^2][sin^2 (x)] } = sqrt { (2a^2)[1+cos(x)] } = [sqrt(2)]a {sqrt [1+cos(x)]} Then s = 2[sqrt(2)]a . Int {sqrt [1+cos(x)]} dx Let 1+cos(x) = 1+2cos^2 (x/2) - 1 = 2cos^2 (x/2) s = 2[sqrt(2)]a . Int {sqrt [2cos^2 (x/2)]} dx s = 4a . Int [cos(x/2)] dx where x vary from 0 to Pi s = 4a [sin(x/2)]/(1/2) s = 8a [sin(Pi/2) - sin(0)] s = 8a
