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sin2xdx/sin^4xdx
Solving integrals/ sin2xdx/sin^4xdx

Integral of sin2xdx/sin^4xdx dx

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∫01sin⁡(2x)1⋅1sin⁡4(x)1 dx\int\limits_{0}^{1} \sin{\left(2 x \right)} 1 \cdot \frac{1}{\sin^{4}{\left(x \right)}} 1\, dx 
Integral(sin(2*x)*1*1/sin(x)^4, (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:

      ∫2sin⁡(x)cos⁡(x)sin⁡4(x) dx=2∫sin⁡(x)cos⁡(x)sin⁡4(x) dx\int \frac{2 \sin{\left(x \right)} \cos{\left(x \right)}}{\sin^{4}{\left(x \right)}}\, dx = 2 \int \frac{\sin{\left(x \right)} \cos{\left(x \right)}}{\sin^{4}{\left(x \right)}}\, dx

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

        Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

        ∫1u3 du\int \frac{1}{u^{3}}\, du

        1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

          ∫1u3 du=−12u2\int \frac{1}{u^{3}}\, du = - \frac{1}{2 u^{2}}

        Now substitute uu back in:

        −12sin⁡2(x)- \frac{1}{2 \sin^{2}{\left(x \right)}}

      So, the result is: −1sin⁡2(x)- \frac{1}{\sin^{2}{\left(x \right)}}

    Method #2

    1. Rewrite the integrand:

      sin⁡(2x)1⋅1sin⁡4(x)1=2cos⁡(x)sin⁡3(x)\sin{\left(2 x \right)} 1 \cdot \frac{1}{\sin^{4}{\left(x \right)}} 1 = \frac{2 \cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}

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

      ∫2cos⁡(x)sin⁡3(x) dx=2∫cos⁡(x)sin⁡3(x) dx\int \frac{2 \cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}\, dx = 2 \int \frac{\cos{\left(x \right)}}{\sin^{3}{\left(x \right)}}\, dx

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

        Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

        ∫1u3 du\int \frac{1}{u^{3}}\, du

        1. The integral of unu^{n} is un+1n+1\frac{u^{n + 1}}{n + 1} when n≠−1n \neq -1:

          ∫1u3 du=−12u2\int \frac{1}{u^{3}}\, du = - \frac{1}{2 u^{2}}

        Now substitute uu back in:

        −12sin⁡2(x)- \frac{1}{2 \sin^{2}{\left(x \right)}}

      So, the result is: −1sin⁡2(x)- \frac{1}{\sin^{2}{\left(x \right)}}

  2. Add the constant of integration:

    −1sin⁡2(x)+constant- \frac{1}{\sin^{2}{\left(x \right)}}+ \mathrm{constant}

The answer is:

−1sin⁡2(x)+constant- \frac{1}{\sin^{2}{\left(x \right)}}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                     
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 |               1                  1   
 | sin(2*x)*1*-------*1 dx = C - -------
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 |            sin (x)            sin (x)
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∫sin⁡(2x)1⋅1sin⁡4(x)1 dx=C−1sin⁡2(x)\int \sin{\left(2 x \right)} 1 \cdot \frac{1}{\sin^{4}{\left(x \right)}} 1\, dx = C - \frac{1}{\sin^{2}{\left(x \right)}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.902000000000000-1000000000000
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The answer [src] 
oo
∞\infty 
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oo
∞\infty
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Numerical answer [src] 
1.83073007580698e+38
1.83073007580698e+38
The graph
Integral of sin2xdx/sin^4xdx dx

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

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