Integrate the function y = x*sin^3x

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neela | High School Teacher | (Level 3) Valedictorian

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We know that Int (u(x)*v(x)) dx = u(x)Int (v(x) - Int {u'(x)*Intv(x)dx}dx

Also we know that sin3x = 3sinx -4sin^3x.

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

We use the above results to integrate  xsin^3x = x(3sinx-sin3x)/4.

u(x) = x and v(x) = (3sinx -sin3x)/4

u'x) = 1 and Int v(x) = Int {(3sinx - sin3x)/4 }dx = {-3cosx - (-cos3x/3)/4} = (-9cosx +cos3x)/12.

Therefore Int x sin^3x dx = Int { x* Int (sinx-sin3x)dx/4 - Int (x'*Int(sinx-sin3xdx) dx

=x(-9cosx+cos3x)/12 - Int (1*-9cosx+cos3xdx/12)

= x(-9cosx+cos3x)/12 +9sinx/12 - sin3x/36

Therefore Int xsin^3x = x(cos3x-9cosx)/12 + (3sinx)/4 - sin3x/36 .

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll 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:

cos 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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