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Solving integrals/ п/0^8cos4xdx

Integral of п/0^8cos4xdx dx

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The solution

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∫01π0cos⁡(4x) dx\int\limits_{0}^{1} \frac{\pi}{0} \cos{\left(4 x \right)}\, dx 
Integral((pi/0)*cos(4*x), (x, 0, 1))
Detail solution

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

    ∫π0cos⁡(4x) dx=∞~∫cos⁡(4x) dx\int \frac{\pi}{0} \cos{\left(4 x \right)}\, dx = \tilde{\infty} \int \cos{\left(4 x \right)}\, dx

    1. Let u=4xu = 4 x.

      Then let du=4dxdu = 4 dx and substitute du4\frac{du}{4}:

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

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

        ∫cos⁡(u) du=∫cos⁡(u) du4\int \cos{\left(u \right)}\, du = \frac{\int \cos{\left(u \right)}\, du}{4}

        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)4\frac{\sin{\left(u \right)}}{4}

      Now substitute uu back in:

      sin⁡(4x)4\frac{\sin{\left(4 x \right)}}{4}

    So, the result is: ∞~sin⁡(4x)\tilde{\infty} \sin{\left(4 x \right)}

  2. Add the constant of integration:

    ∞~sin⁡(4x)+constant\tilde{\infty} \sin{\left(4 x \right)}+ \mathrm{constant}

The answer is:

∞~sin⁡(4x)+constant\tilde{\infty} \sin{\left(4 x \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
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 | --*cos(4*x) dx = C + zoo*sin(4*x)
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∫π0cos⁡(4x) dx=C+∞~sin⁡(4x)\int \frac{\pi}{0} \cos{\left(4 x \right)}\, dx = C + \tilde{\infty} \sin{\left(4 x \right)}
Simplify
The graph
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The answer [src] 
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zoo* |  cos(4*x) dx
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∞~∫01cos⁡(4x) dx\tilde{\infty} \int\limits_{0}^{1} \cos{\left(4 x \right)}\, dx 
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zoo* |  cos(4*x) dx
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∞~∫01cos⁡(4x) dx\tilde{\infty} \int\limits_{0}^{1} \cos{\left(4 x \right)}\, dx 
±oo*Integral(cos(4*x), (x, 0, 1))

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

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