# Evaluate the integral? : #int3xsinxcosx dx#?

# int \ 3xsinxcosx \ dx = 3/8 sin2x -3/4xcos2x + C #

We wish to assess:

Firstly take note that when utilizing the sine function's double angle formula:

The integral can be expressed as:

Essentially, we would like to identify one function that simplifies when differentiated and another that simplifies when integrated (or is at least integrable). To do this, we can use Integration By Parts (IBP).

After that, entering the IBP formula:

We get:

Hence:

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To evaluate the integral ( \int 3x \sin(x) \cos(x) , dx ), you can use integration by parts. Let's denote:

[ u = 3x ] [ dv = \sin(x) \cos(x) , dx ]

Then, differentiate ( u ) to get ( du ) and integrate ( dv ) to get ( v ):

[ du = 3 , dx ] [ v = \int \sin(x) \cos(x) , dx ]

Now, integrate by parts using the formula:

[ \int u , dv = uv - \int v , du ]

Substitute the values of ( u ), ( dv ), ( du ), and ( v ) into the formula:

[ = 3x \left( \int \sin(x) \cos(x) , dx \right) - \int \left( \int \sin(x) \cos(x) , dx \right) 3 , dx ]

Now, you need to evaluate the integrals:

[ v = \int \sin(x) \cos(x) , dx ]

[ = -\frac{1}{2} \sin^2(x) + C ]

[ \int \left( \int \sin(x) \cos(x) , dx \right) 3 , dx ]

[ = \int \left( -\frac{1}{2} \sin^2(x) \right) 3 , dx ]

[ = -\frac{3}{2} \int \sin^2(x) , dx ]

Now, you can use the double angle identity ( \sin^2(x) = \frac{1 - \cos(2x)}{2} ):

[ = -\frac{3}{2} \int \left( \frac{1 - \cos(2x)}{2} \right) , dx ]

[ = -\frac{3}{4} \int (1 - \cos(2x)) , dx ]

[ = -\frac{3}{4} \left( x - \frac{1}{2} \sin(2x) \right) + C ]

Now, substitute these results back into the integration by parts formula:

[ = 3x \left( -\frac{1}{2} \sin^2(x) \right) + \frac{3}{4} \left( x - \frac{1}{2} \sin(2x) \right) + C ]

[ = -\frac{3}{2}x\sin^2(x) + \frac{3}{4}x - \frac{3}{8}\sin(2x) + C ]

Therefore, the solution to the integral ( \int 3x \sin(x) \cos(x) , dx ) is ( -\frac{3}{2}x\sin^2(x) + \frac{3}{4}x - \frac{3}{8}\sin(2x) + C ), where ( C ) is the constant of integration.

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