Calculate the indefinite integral∫3xsinx(4-x^2)dx Would like an explanation and steps, not just an answer please. Thank you in advance

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to expand the integrand such that:

`int 3x*sinx*(4-x^2)dx = 3 int (4xsin x - x^3sin x) dx`  

Using the linearity yields:

`int 3x*sinx*(4-x^2)dx = 3 int 4xsin x dx - 3 int x^3sin x dx`

You need to integrate by parts using the following formula such that:

`int udv = uv - int vdu`

Reasoning by analogy yields:

`3 int 4xsin x dx =12 int xsin x dx`

`u = x => du = dx`

`dv = sin x dx => v = -cos x`

`12 int xsin x dx = 12(-xcos x + int cos x dx) `

`12 int xsin x dx = -12x cos x - 12 sin x + ` c

You need to use integration by parts to solve `3 int x^3sin x dx`  such that:

`u = x^3 => du = 3x^2 dx`

`dv = sin x dx => v = - cos x`

`3 int x^3sin x dx = 3(-x^3 cos x + 3int x^2 cos x dx)`

`3 int x^3sin x dx = -3x^3 cos x + 9int x^2 cos x dx`

You need to use parts to evaluate `int x^2 cos x dx`  such that:

`u = x^2 => du = 2x dx`

`dv = cos x dx => v = sin x`

`int x^2 cos x dx = x^2 sin x - int 2x sin x dx`

`int x^2 cos x dx = x^2 sin x - 2int x sin x dx`

You need to use parts to evaluate `int x sin x dx`  such that:

`u = x => du = dx`

`dv = sin x dx => v = - cos x`

`int x sin x dx = -x cos x + int cos x dx`

`int x sin x dx = -x cos x + sin x + c`

`int x^2 cos x dx = x^2 sin x - 2(-x cos x + sin x)`

`int x^2 cos x dx = x^2 sin x + 2x cos x - 2 sin x + c`

`3 int x^3sin x dx = -3x^3 cos x + 9x^2 sin x + 18x cos x - 18 sin x + c`

`int 3x*sinx*(4-x^2)dx =-12x cos x - 12 sin x - (-3x^3 cos x + 9x^2 sin x + 18x cos x - 18 sin x) + c`

`int 3x*sinx*(4-x^2)dx = -12x cos x - 12 sin x + 3x^3 cos x - 9x^2 sin x - 18x cos x- 18 sin x + c`

`int 3x*sinx*(4-x^2)dx = 3x^3 cos x - 9x^2 sin x - 30x cos x -30sin x + c`

Hence, evaluating the given indefinite integral, using integration by parts, yields `int 3x*sinx*(4-x^2)dx = 3x^3 cos x - 9x^2 sin x - 30x cos x -30 sin x + c.`

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quantatanu | Student, Undergraduate | (Level 1) Valedictorian

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The integral is:

∫ 3 x sin(x) (4-x^2) dx

= 3 ∫ [4 x Sin(x) - x^3 Sin(x) ] dx  

= 12 ∫ x Sin(x) dx  - 3 ∫ x^3 Sin(x) dx

Using integration by parts for the integral:

x^2 Cos(x), 

∫ f dg = fg - ∫ g df

where,

f = x^2, dg = Cos(x) dx,  df = 2x dx, g = Sin(x) so,

so the integration becomes,

= 3 x^3 Cos(x) - 9 ∫ x^2 Cos(x) dx + 12 ∫ x Sin(x) dx

= 3 x^3 Cos(x) - 9 x^2 Sin(x) + 18 ∫ x Sin(x) dx  

                                                  + 12 ∫ x Sin(x) dx

 

Again using integration by parts for x Sin(x) dx :

Now:  f =x, dg = Sin(x) dx, df = dx, g = -Cos(x)

So the integration proceeds as:

= 3 x^3 Cos(x) - 9 x^2 Sin(x) + 18 Sin(x) - 30 x Cos(x)

   + 12 ∫ Cos(x) dx

= 3 x^3 Cos(x) - 9 x^2 Sin(x) + 30 Sin(x) - 30 x Cos(x)

 

Answer is:

3 x^3 Cos(x) - 9 x^2 Sin(x) + 30 Sin(x) - 30 x Cos(x)

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