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Solving integrals/ 1+sin2x/sinx^2

Integral of 1+sin2x/sinx^2 dx

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The solution

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

  1. Integrate term-by-term:

    1. The integral of a constant is the constant times the variable of integration:

      ∫1 dx=x\int 1\, dx = x

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

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

          Then let du=−2cos⁡(x)dxsin⁡3(x)du = - \frac{2 \cos{\left(x \right)} dx}{\sin^{3}{\left(x \right)}} and substitute −du2- \frac{du}{2}:

          ∫14u du\int \frac{1}{4 u}\, du

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

            ∫(−12u) du=−∫1u du2\int \left(- \frac{1}{2 u}\right)\, du = - \frac{\int \frac{1}{u}\, du}{2}

            1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

            So, the result is: −log⁡(u)2- \frac{\log{\left(u \right)}}{2}

          Now substitute uu back in:

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

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

      Method #2

      1. Rewrite the integrand:

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

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

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

          ∫1u du\int \frac{1}{u}\, du

          1. The integral of 1u\frac{1}{u} is log⁡(u)\log{\left(u \right)}.

          Now substitute uu back in:

          log⁡(sin⁡(x))\log{\left(\sin{\left(x \right)} \right)}

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

    The result is: x−log⁡(1sin⁡2(x))x - \log{\left(\frac{1}{\sin^{2}{\left(x \right)}} \right)}

  2. Add the constant of integration:

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

The answer is:

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

The answer (Indefinite) [src] 
  /                                        
 |                                         
 | /    sin(2*x)\                 /   1   \
 | |1 + --------| dx = C + x - log|-------|
 | |       2    |                 |   2   |
 | \    sin (x) /                 \sin (x)/
 |                                         
/
∫(1+sin⁡(2x)sin⁡2(x)) dx=C+x−log⁡(1sin⁡2(x))\int \left(1 + \frac{\sin{\left(2 x \right)}}{\sin^{2}{\left(x \right)}}\right)\, dx = C + x - \log{\left(\frac{1}{\sin^{2}{\left(x \right)}} \right)}
Simplify
The answer [src] 
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∞\infty 
=
= 
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∞\infty
Упростить
Numerical answer [src] 
88.8356847754476
88.8356847754476

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

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