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Solving integrals/ pi(3sin4x)^2

Integral of pi(3sin4x)^2 dx

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The solution

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∫01π(3sin⁡(4x))2 dx\int\limits_{0}^{1} \pi \left(3 \sin{\left(4 x \right)}\right)^{2}\, dx 
Integral(pi*(3*sin(4*x))^2, (x, 0, 1))
Detail solution

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

    ∫π(3sin⁡(4x))2 dx=π∫(3sin⁡(4x))2 dx\int \pi \left(3 \sin{\left(4 x \right)}\right)^{2}\, dx = \pi \int \left(3 \sin{\left(4 x \right)}\right)^{2}\, dx

    1. Don't know the steps in finding this integral.

      But the integral is

      9xsin⁡2(4x)2+9xcos⁡2(4x)2−9sin⁡(4x)cos⁡(4x)8\frac{9 x \sin^{2}{\left(4 x \right)}}{2} + \frac{9 x \cos^{2}{\left(4 x \right)}}{2} - \frac{9 \sin{\left(4 x \right)} \cos{\left(4 x \right)}}{8}

    So, the result is: π(9xsin⁡2(4x)2+9xcos⁡2(4x)2−9sin⁡(4x)cos⁡(4x)8)\pi \left(\frac{9 x \sin^{2}{\left(4 x \right)}}{2} + \frac{9 x \cos^{2}{\left(4 x \right)}}{2} - \frac{9 \sin{\left(4 x \right)} \cos{\left(4 x \right)}}{8}\right)

  2. Now simplify:

    9π(4xsin⁡2(4x)+4xcos⁡2(4x)−sin⁡(4x)cos⁡(4x))8\frac{9 \pi \left(4 x \sin^{2}{\left(4 x \right)} + 4 x \cos^{2}{\left(4 x \right)} - \sin{\left(4 x \right)} \cos{\left(4 x \right)}\right)}{8}

  3. Add the constant of integration:

    9π(4xsin⁡2(4x)+4xcos⁡2(4x)−sin⁡(4x)cos⁡(4x))8+constant\frac{9 \pi \left(4 x \sin^{2}{\left(4 x \right)} + 4 x \cos^{2}{\left(4 x \right)} - \sin{\left(4 x \right)} \cos{\left(4 x \right)}\right)}{8}+ \mathrm{constant}

The answer is:

9π(4xsin⁡2(4x)+4xcos⁡2(4x)−sin⁡(4x)cos⁡(4x))8+constant\frac{9 \pi \left(4 x \sin^{2}{\left(4 x \right)} + 4 x \cos^{2}{\left(4 x \right)} - \sin{\left(4 x \right)} \cos{\left(4 x \right)}\right)}{8}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                                                    
 |                              /                               2               2     \
 |                2             |  9*cos(4*x)*sin(4*x)   9*x*cos (4*x)   9*x*sin (4*x)|
 | pi*(3*sin(4*x))  dx = C + pi*|- ------------------- + ------------- + -------------|
 |                              \           8                  2               2      /
/
∫π(3sin⁡(4x))2 dx=C+π(9xsin⁡2(4x)2+9xcos⁡2(4x)2−9sin⁡(4x)cos⁡(4x)8)\int \pi \left(3 \sin{\left(4 x \right)}\right)^{2}\, dx = C + \pi \left(\frac{9 x \sin^{2}{\left(4 x \right)}}{2} + \frac{9 x \cos^{2}{\left(4 x \right)}}{2} - \frac{9 \sin{\left(4 x \right)} \cos{\left(4 x \right)}}{8}\right)
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.90050
Plot the graph f(x)
The answer [src] 
     /1   cos(4)*sin(4)\
9*pi*|- - -------------|
     \2         8      /
9π(−sin⁡(4)cos⁡(4)8+12)9 \pi \left(- \frac{\sin{\left(4 \right)} \cos{\left(4 \right)}}{8} + \frac{1}{2}\right) 
=
= 
     /1   cos(4)*sin(4)\
9*pi*|- - -------------|
     \2         8      /
9π(−sin⁡(4)cos⁡(4)8+12)9 \pi \left(- \frac{\sin{\left(4 \right)} \cos{\left(4 \right)}}{8} + \frac{1}{2}\right) 
9*pi*(1/2 - cos(4)*sin(4)/8)
Numerical answer [src] 
12.3888266040138
12.3888266040138

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

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