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Find the antiderivative of the function f(x)=x*sin^3x.

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fab10 | Student, Undergraduate | eNoter

Posted November 13, 2010 at 1:59 AM via web

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Find the antiderivative of the function f(x)=x*sin^3x.

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giorgiana1976 | College Teacher | Valedictorian

Posted November 13, 2010 at 1:59 AM (Answer #1)

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To determine the antiderivative of the given function, we'll have to integrate the given function by parts, using the formula:

Int f*g'  = f*g - Int f'*g

We'll put f = x => f' = 1

We'll put g' = (sinx)^3 and we'll have to integrate (sinx)^3 to find the function g.

We'll write the function as a product:

(sinx)^3 = (sinx)^2*sin x

We'll integrate both sides:

Int (sinx)^3dx = Int [(sinx)^2*sin x]dx

We'll write (sinx)^2  = 1 - (cosx)^2

Int [(sinx)^2*sin x]dx = Int [(1 - (cosx)^2)*sin x]dx

We'll remove the brackets:

Int [(1 - (cosx)^2)*sin x]dx  = Int sin xdx - Int (cosx)^2*sin xdx

We'll solve Int (cosx)^2*sin xdx using substitution technique:

cos x = t

We'll differentiate both sides:

-sin xdx = dt

We'll re-write the integral, changing the variable:

Int (cosx)^2*sin xdx = Int t^2dt

Int t^2dt = t^3/3 + C

Int (cosx)^2*sin xdx = (cos x)^3/3 + C

Int (sinx)^3dx = Int sin xdx - Int (cosx)^2*sin xdx

Int (sinx)^3dx = -cos x - (cos x)^3/3 + C

So, if g' = (sinx)^3 => g = -cos x - (cos x)^3/3

Now, we can integrate by parts:

Int x*(sinx)^3dx = x*[-cos x - (cos x)^3/3] - Int [ -cos x - (cos x)^3/3] dx

We'll apply the additive property of integrals:

Int x*(sinx)^3dx = x*[-cos x - (cos x)^3/3] + Int cos x dx + (1/3)Int (cos x)^3 dx

We'll write the function as a product:

(cos x)^3 = (cos x)^2*cos x

We'll integrate both sides:

Int (cos x)^3 dx = Int (cos x)^2*cos x dx

We'll write (cos x)^2  = 1 - (sin x)^2

Int (cos x)^2*cos x dx = Int [(1 - (sin x)^2)*cos x]dx

We'll remove the brackets:

Int [(1 - (sin x)^2)*cos x]dx  = Int cos xdx - Int (sinx)^2*cos xdx

We'll solve Int (sinx)^2*cos xdx using substitution technique:

sin x = t

We'll differentiate both sides:

cos xdx = dt

We'll re-write the integral, changing the variable:

Int (sinx)^2*cos xdx = Int t^2dt

Int t^2dt = t^3/3 + C

Int [(1 - (sin x)^2)*cos x]dx  = sin x - (sin x)^3/3 + C

Int x*(sinx)^3dx = x*[-cos x - (cos x)^3/3] +sinx + (1/3)[sin x - (sin x)^3/3] + C

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neela | High School Teacher | Valedictorian

Posted November 13, 2010 at 2:35 AM (Answer #2)

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To find the antiderivative or integral of f(x) = xsin^3x.

We know that sin 3x = 3sinx-4sin^3x.

Therefore sin^3x = (3sinx -sin3x)/4

Therefore x sin^3x = x(3sinx-sin3x)/4....(1)

Int xsinx dx = x Int sinx dx - Int{x' int sinx dx) dx

Int xsinx dx = x (-cosx) +Int cosx dx = -xcosx + sinx +const

Therefore (3/4) Int xsinx = - 3/4xcosx +3/4 sinx +C1..........(2).

Int xsin3x dx = x (-cos3x/3) + (1/3)Int {x' cos3x }dx

Int x sin3x dx =  - (1/3) xcos3x + (1/9) sin3x + a Const

Therefore - 1/4 Int x sin3x = 1/12xcos3x - 1/36 sin3x............(3).

Therefore adding (3) and (4) we get:

 Int x(3sinx- sin3x)/4 dx = {-(3/4) xcosx +3/4sinx +1/12xcos3x -1/36sin3x} + C

Therefore Int xsin^3x = {-(3/4) xcosx +3/4sinx +1/12xcos3x -1/36sin3x} + C.

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