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Solving integrals/ cos(2*pi*x)f(t)

Integral of cos(2*pi*x)f(t) dy

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∫01ftcos⁡(2πx) dx\int\limits_{0}^{1} f t \cos{\left(2 \pi x \right)}\, dx 
Integral(cos(2*pi*x)*f*t, (x, 0, 1))
Detail solution

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

    ∫ftcos⁡(2πx) dx=ft∫cos⁡(2πx) dx\int f t \cos{\left(2 \pi x \right)}\, dx = f t \int \cos{\left(2 \pi x \right)}\, dx

    1. Let u=2πxu = 2 \pi x.

      Then let du=2πdxdu = 2 \pi dx and substitute du2π\frac{du}{2 \pi}:

      ∫cos⁡(u)4π2 du\int \frac{\cos{\left(u \right)}}{4 \pi^{2}}\, du

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

        ∫cos⁡(u)2π du=∫cos⁡(u) du2π\int \frac{\cos{\left(u \right)}}{2 \pi}\, du = \frac{\int \cos{\left(u \right)}\, du}{2 \pi}

        1. The integral of cosine is sine:

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

        So, the result is: sin⁡(u)2π\frac{\sin{\left(u \right)}}{2 \pi}

      Now substitute uu back in:

      sin⁡(2πx)2π\frac{\sin{\left(2 \pi x \right)}}{2 \pi}

    So, the result is: ftsin⁡(2πx)2π\frac{f t \sin{\left(2 \pi x \right)}}{2 \pi}

  2. Add the constant of integration:

    ftsin⁡(2πx)2π+constant\frac{f t \sin{\left(2 \pi x \right)}}{2 \pi}+ \mathrm{constant}

The answer is:

ftsin⁡(2πx)2π+constant\frac{f t \sin{\left(2 \pi x \right)}}{2 \pi}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                        
 |                          f*t*sin(2*pi*x)
 | cos(2*pi*x)*f*t dx = C + ---------------
 |                                2*pi     
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f t sin⁡(2 π x)2 π{{f\,t\,\sin \left(2\,\pi\,x\right)}\over{2\,\pi}}
Simplify
The answer [src] 
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f sin⁡(2 π) t2 π{{f\,\sin \left(2\,\pi\right)\,t}\over{2\,\pi}} 
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    Use the examples entering the upper and lower limits of integration.

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