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Solving integrals/ exp(sin^2x)*sin2xdx

Integral of exp(sin^2x)*sin2xdx dx

Limits of integration:

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The solution

You have entered [src] 
  1                     
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 |  e       *sin(2*x) dx
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∫01esin⁡2(x)sin⁡(2x) dx\int\limits_{0}^{1} e^{\sin^{2}{\left(x \right)}} \sin{\left(2 x \right)}\, dx 
Integral(exp(sin(x)^2)*sin(2*x), (x, 0, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

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

      ∫2esin⁡2(x)sin⁡(x)cos⁡(x) dx=2∫esin⁡2(x)sin⁡(x)cos⁡(x) dx\int 2 e^{\sin^{2}{\left(x \right)}} \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int e^{\sin^{2}{\left(x \right)}} \sin{\left(x \right)} \cos{\left(x \right)}\, dx

      1. Let u=sin⁡2(x)u = \sin^{2}{\left(x \right)}.

        Then let du=2sin⁡(x)cos⁡(x)dxdu = 2 \sin{\left(x \right)} \cos{\left(x \right)} dx and substitute du2\frac{du}{2}:

        ∫eu2 du\int \frac{e^{u}}{2}\, du

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

          False\text{False}

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          So, the result is: eu2\frac{e^{u}}{2}

        Now substitute uu back in:

        esin⁡2(x)2\frac{e^{\sin^{2}{\left(x \right)}}}{2}

      So, the result is: esin⁡2(x)e^{\sin^{2}{\left(x \right)}}

    Method #2

    1. Rewrite the integrand:

      esin⁡2(x)sin⁡(2x)=2esin⁡2(x)sin⁡(x)cos⁡(x)e^{\sin^{2}{\left(x \right)}} \sin{\left(2 x \right)} = 2 e^{\sin^{2}{\left(x \right)}} \sin{\left(x \right)} \cos{\left(x \right)}

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

      ∫2esin⁡2(x)sin⁡(x)cos⁡(x) dx=2∫esin⁡2(x)sin⁡(x)cos⁡(x) dx\int 2 e^{\sin^{2}{\left(x \right)}} \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int e^{\sin^{2}{\left(x \right)}} \sin{\left(x \right)} \cos{\left(x \right)}\, dx

      1. Let u=sin⁡2(x)u = \sin^{2}{\left(x \right)}.

        Then let du=2sin⁡(x)cos⁡(x)dxdu = 2 \sin{\left(x \right)} \cos{\left(x \right)} dx and substitute du2\frac{du}{2}:

        ∫eu2 du\int \frac{e^{u}}{2}\, du

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

          False\text{False}

          1. The integral of the exponential function is itself.

            ∫eu du=eu\int e^{u}\, du = e^{u}

          So, the result is: eu2\frac{e^{u}}{2}

        Now substitute uu back in:

        esin⁡2(x)2\frac{e^{\sin^{2}{\left(x \right)}}}{2}

      So, the result is: esin⁡2(x)e^{\sin^{2}{\left(x \right)}}

  2. Add the constant of integration:

    esin⁡2(x)+constante^{\sin^{2}{\left(x \right)}}+ \mathrm{constant}

The answer is:

esin⁡2(x)+constante^{\sin^{2}{\left(x \right)}}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                   
 |                                    
 |     2                          2   
 |  sin (x)                    sin (x)
 | e       *sin(2*x) dx = C + e       
 |                                    
/
∫esin⁡2(x)sin⁡(2x) dx=C+esin⁡2(x)\int e^{\sin^{2}{\left(x \right)}} \sin{\left(2 x \right)}\, dx = C + e^{\sin^{2}{\left(x \right)}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.9004
Plot the graph f(x)
The answer [src] 
         2   
      sin (1)
-1 + e
−1+esin⁡2(1)-1 + e^{\sin^{2}{\left(1 \right)}} 
=
= 
         2   
      sin (1)
-1 + e
−1+esin⁡2(1)-1 + e^{\sin^{2}{\left(1 \right)}} 
-1 + exp(sin(1)^2)
Numerical answer [src] 
1.03007638063326
1.03007638063326

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

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