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Solving integrals/ sin(x+y)

Integral of 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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∫01sin⁡(x+y) dx\int\limits_{0}^{1} \sin{\left(x + y \right)}\, dx 
Integral(sin(x + y), (x, 0, 1))
Detail solution

  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)}

  2. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
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 | sin(x + y) dx = C - cos(x + y)
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∫sin⁡(x+y) dx=C−cos⁡(x+y)\int \sin{\left(x + y \right)}\, dx = C - \cos{\left(x + y \right)}
Simplify
The answer [src] 
-cos(1 + y) + cos(y)
cos⁡(y)−cos⁡(y+1)\cos{\left(y \right)} - \cos{\left(y + 1 \right)} 
=
= 
-cos(1 + y) + cos(y)
cos⁡(y)−cos⁡(y+1)\cos{\left(y \right)} - \cos{\left(y + 1 \right)} 
-cos(1 + y) + cos(y)

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

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