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sinx/(2cosx)/2
Solving integrals/ sinx/(2cosx)/2

Integral of sinx/(2cosx)/2 dx

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

    ∫sin⁡(x)2⋅2cos⁡(x) dx=∫sin⁡(x)2cos⁡(x) dx2\int \frac{\sin{\left(x \right)}}{2 \cdot 2 \cos{\left(x \right)}}\, dx = \frac{\int \frac{\sin{\left(x \right)}}{2 \cos{\left(x \right)}}\, dx}{2}

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

      Then let du=−2sin⁡(x)dxdu = - 2 \sin{\left(x \right)} dx 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⁡(2cos⁡(x))2- \frac{\log{\left(2 \cos{\left(x \right)} \right)}}{2}

    So, the result is: −log⁡(2cos⁡(x))4- \frac{\log{\left(2 \cos{\left(x \right)} \right)}}{4}

  2. Add the constant of integration:

    −log⁡(2cos⁡(x))4+constant- \frac{\log{\left(2 \cos{\left(x \right)} \right)}}{4}+ \mathrm{constant}

The answer is:

−log⁡(2cos⁡(x))4+constant- \frac{\log{\left(2 \cos{\left(x \right)} \right)}}{4}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                 
 |                                  
 |   sin(x)            log(2*cos(x))
 | ---------- dx = C - -------------
 | 2*cos(x)*2                4      
 |                                  
/
−log⁡cos⁡x4-{{\log \cos x}\over{4}}
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.900.00.5
Plot the graph f(x)
The answer [src] 
-log(cos(1)) 
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      4
−log⁡cos⁡14-{{\log \cos 1}\over{4}} 
=
= 
-log(cos(1)) 
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      4
−log⁡(cos⁡(1))4- \frac{\log{\left(\cos{\left(1 \right)} \right)}}{4}
Упростить
Numerical answer [src] 
0.153906617596504
0.153906617596504
The graph
Integral of sinx/(2cosx)/2 dx

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

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