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Solving integrals/ x*(sin(x+y))

Integral of x*(sin(x+y)) dx

Limits of integration:

from to
v

The graph:

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Piecewise:

The solution

You have entered [src] 
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∫0pxsin⁡(x+y) dx\int\limits_{0}^{p} x \sin{\left(x + y \right)}\, dx 
Integral(x*sin(x + y), (x, 0, p))
Detail solution

  1. Use integration by parts:

    ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

    Let u(x)=xu{\left(x \right)} = x and let dv⁡(x)=sin⁡(x+y)\operatorname{dv}{\left(x \right)} = \sin{\left(x + y \right)}.

    Then du⁡(x)=1\operatorname{du}{\left(x \right)} = 1.

    To find v(x)v{\left(x \right)}:

    1. Let u=x+yu = x + y.

      Then let du=dxdu = dx and substitute dudu:

      ∫sin⁡(u) du\int \sin{\left(u \right)}\, du

      1. The integral of sine is negative cosine:

        ∫sin⁡(u) du=−cos⁡(u)\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}

      Now substitute uu back in:

      −cos⁡(x+y)- \cos{\left(x + y \right)}

    Now evaluate the sub-integral.

  2. The integral of a constant times a function is the constant times the integral of the function:

    ∫(−cos⁡(x+y)) dx=−∫cos⁡(x+y) dx\int \left(- \cos{\left(x + y \right)}\right)\, dx = - \int \cos{\left(x + y \right)}\, dx

    1. Let u=x+yu = x + y.

      Then let du=dxdu = dx and substitute dudu:

      ∫cos⁡(u) du\int \cos{\left(u \right)}\, du

      1. The integral of cosine is sine:

        ∫cos⁡(u) du=sin⁡(u)\int \cos{\left(u \right)}\, du = \sin{\left(u \right)}

      Now substitute uu back in:

      sin⁡(x+y)\sin{\left(x + y \right)}

    So, the result is: −sin⁡(x+y)- \sin{\left(x + y \right)}

  3. Add the constant of integration:

    −xcos⁡(x+y)+sin⁡(x+y)+constant- x \cos{\left(x + y \right)} + \sin{\left(x + y \right)}+ \mathrm{constant}

The answer is:

−xcos⁡(x+y)+sin⁡(x+y)+constant- x \cos{\left(x + y \right)} + \sin{\left(x + y \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
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 | x*sin(x + y) dx = C - x*cos(x + y) + sin(x + y)
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∫xsin⁡(x+y) dx=C−xcos⁡(x+y)+sin⁡(x+y)\int x \sin{\left(x + y \right)}\, dx = C - x \cos{\left(x + y \right)} + \sin{\left(x + y \right)}
Simplify
The answer [src] 
-sin(y) - p*cos(p + y) + sin(p + y)
−pcos⁡(p+y)−sin⁡(y)+sin⁡(p+y)- p \cos{\left(p + y \right)} - \sin{\left(y \right)} + \sin{\left(p + y \right)} 
=
= 
-sin(y) - p*cos(p + y) + sin(p + y)
−pcos⁡(p+y)−sin⁡(y)+sin⁡(p+y)- p \cos{\left(p + y \right)} - \sin{\left(y \right)} + \sin{\left(p + y \right)} 
-sin(y) - p*cos(p + y) + sin(p + y)

    Use the examples entering the upper and lower limits of integration.

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