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t^3sin(2t^4)dt

Integral of t^3sin(2t^4)dt dx

Limits of integration:

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The graph:

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Piecewise:

The solution

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01t3sin(2t4)1dt\int\limits_{0}^{1} t^{3} \sin{\left(2 t^{4} \right)} 1\, dt
Integral(t^3*sin(2*t^4)*1, (t, 0, 1))
Detail solution
  1. Let u=2t4u = 2 t^{4}.

    Then let du=8t3dtdu = 8 t^{3} dt and substitute du8\frac{du}{8}:

    sin(u)64du\int \frac{\sin{\left(u \right)}}{64}\, du

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

      sin(u)8du=sin(u)du8\int \frac{\sin{\left(u \right)}}{8}\, du = \frac{\int \sin{\left(u \right)}\, du}{8}

      1. The integral of sine is negative cosine:

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

      So, the result is: cos(u)8- \frac{\cos{\left(u \right)}}{8}

    Now substitute uu back in:

    cos(2t4)8- \frac{\cos{\left(2 t^{4} \right)}}{8}

  2. Add the constant of integration:

    cos(2t4)8+constant- \frac{\cos{\left(2 t^{4} \right)}}{8}+ \mathrm{constant}


The answer is:

cos(2t4)8+constant- \frac{\cos{\left(2 t^{4} \right)}}{8}+ \mathrm{constant}

The answer (Indefinite) [src]
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 |  3    /   4\            cos\2*t /
 | t *sin\2*t /*1 dt = C - ---------
 |                             8    
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cos(2t4)8-{{\cos \left(2\,t^4\right)}\over{8}}
The graph
0.001.000.100.200.300.400.500.600.700.800.901-1
The answer [src]
1   cos(2)
- - ------
8     8   
18cos28{{1}\over{8}}-{{\cos 2}\over{8}}
=
=
1   cos(2)
- - ------
8     8   
cos(2)8+18- \frac{\cos{\left(2 \right)}}{8} + \frac{1}{8}
Numerical answer [src]
0.177018354568393
0.177018354568393
The graph
Integral of t^3sin(2t^4)dt dx

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