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cos(1-3x)
Solving integrals/ cos(1-3x)

Integral of cos(1-3x) dx

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∫00cos⁡(1−3x) dx\int\limits_{0}^{0} \cos{\left(1 - 3 x \right)}\, dx 
Integral(cos(1 - 3*x), (x, 0, 0))
Detail solution

  1. Let u=1−3xu = 1 - 3 x.

    Then let du=−3dxdu = - 3 dx and substitute −du3- \frac{du}{3}:

    ∫(−cos⁡(u)3) du\int \left(- \frac{\cos{\left(u \right)}}{3}\right)\, du

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

      ∫cos⁡(u) du=−∫cos⁡(u) du3\int \cos{\left(u \right)}\, du = - \frac{\int \cos{\left(u \right)}\, du}{3}

      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)3- \frac{\sin{\left(u \right)}}{3}

    Now substitute uu back in:

    sin⁡(3x−1)3\frac{\sin{\left(3 x - 1 \right)}}{3}

  2. Add the constant of integration:

    sin⁡(3x−1)3+constant\frac{\sin{\left(3 x - 1 \right)}}{3}+ \mathrm{constant}

The answer is:

sin⁡(3x−1)3+constant\frac{\sin{\left(3 x - 1 \right)}}{3}+ \mathrm{constant}

The answer (Indefinite) [src] 
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 |                       sin(-1 + 3*x)
 | cos(1 - 3*x) dx = C + -------------
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∫cos⁡(1−3x) dx=C+sin⁡(3x−1)3\int \cos{\left(1 - 3 x \right)}\, dx = C + \frac{\sin{\left(3 x - 1 \right)}}{3}
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Integral of cos(1-3x) dx

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

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