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Solving integrals/ 0,5*sin(x+y^2)

Integral of 0,5*sin(x+y^2) dx

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The solution

You have entered [src] 
  1               
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∫01sin⁡(x+y2)2 dx\int\limits_{0}^{1} \frac{\sin{\left(x + y^{2} \right)}}{2}\, dx 
Integral(sin(x + y^2)/2, (x, 0, 1))
Detail solution

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

    ∫sin⁡(x+y2)2 dx=∫sin⁡(x+y2) dx2\int \frac{\sin{\left(x + y^{2} \right)}}{2}\, dx = \frac{\int \sin{\left(x + y^{2} \right)}\, dx}{2}

    1. Let u=x+y2u = x + y^{2}.

      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+y2)- \cos{\left(x + y^{2} \right)}

    So, the result is: −cos⁡(x+y2)2- \frac{\cos{\left(x + y^{2} \right)}}{2}

  2. Add the constant of integration:

    −cos⁡(x+y2)2+constant- \frac{\cos{\left(x + y^{2} \right)}}{2}+ \mathrm{constant}

The answer is:

−cos⁡(x+y2)2+constant- \frac{\cos{\left(x + y^{2} \right)}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                
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 |    /     2\             /     2\
 | sin\x + y /          cos\x + y /
 | ----------- dx = C - -----------
 |      2                    2     
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/
∫sin⁡(x+y2)2 dx=C−cos⁡(x+y2)2\int \frac{\sin{\left(x + y^{2} \right)}}{2}\, dx = C - \frac{\cos{\left(x + y^{2} \right)}}{2}
Simplify
The answer [src] 
   / 2\      /     2\
cos\y /   cos\1 + y /
------- - -----------
   2           2
cos⁡(y2)2−cos⁡(y2+1)2\frac{\cos{\left(y^{2} \right)}}{2} - \frac{\cos{\left(y^{2} + 1 \right)}}{2} 
=
= 
   / 2\      /     2\
cos\y /   cos\1 + y /
------- - -----------
   2           2
cos⁡(y2)2−cos⁡(y2+1)2\frac{\cos{\left(y^{2} \right)}}{2} - \frac{\cos{\left(y^{2} + 1 \right)}}{2} 
cos(y^2)/2 - cos(1 + y^2)/2

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

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