Mister Exam

Other calculators

Solving integrals/ (sin(x/4))^3

Integral of (sin(x/4))^3 dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
 2*p          
  /           
 |            
 |     3/x\   
 |  sin |-| dx
 |      \4/   
 |            
/             
0
∫02psin⁡3(x4) dx\int\limits_{0}^{2 p} \sin^{3}{\left(\frac{x}{4} \right)}\, dx 
Integral(sin(x/4)^3, (x, 0, 2*p))
Detail solution

  1. Rewrite the integrand:

    sin⁡3(x4)=(1−cos⁡2(x4))sin⁡(x4)\sin^{3}{\left(\frac{x}{4} \right)} = \left(1 - \cos^{2}{\left(\frac{x}{4} \right)}\right) \sin{\left(\frac{x}{4} \right)}

  2. There are multiple ways to do this integral.

    Method #1

    1. Let u=cos⁡(x4)u = \cos{\left(\frac{x}{4} \right)}.

      Then let du=−sin⁡(x4)dx4du = - \frac{\sin{\left(\frac{x}{4} \right)} dx}{4} and substitute dudu:

      ∫(4u2−4) du\int \left(4 u^{2} - 4\right)\, du

      1. Integrate term-by-term:

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

          ∫4u2 du=4∫u2 du\int 4 u^{2}\, du = 4 \int u^{2}\, du

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

            ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

          So, the result is: 4u33\frac{4 u^{3}}{3}

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

          ∫(−4) du=−4u\int \left(-4\right)\, du = - 4 u

        The result is: 4u33−4u\frac{4 u^{3}}{3} - 4 u

      Now substitute uu back in:

      4cos⁡3(x4)3−4cos⁡(x4)\frac{4 \cos^{3}{\left(\frac{x}{4} \right)}}{3} - 4 \cos{\left(\frac{x}{4} \right)}

    Method #2

    1. Rewrite the integrand:

      (1−cos⁡2(x4))sin⁡(x4)=−sin⁡(x4)cos⁡2(x4)+sin⁡(x4)\left(1 - \cos^{2}{\left(\frac{x}{4} \right)}\right) \sin{\left(\frac{x}{4} \right)} = - \sin{\left(\frac{x}{4} \right)} \cos^{2}{\left(\frac{x}{4} \right)} + \sin{\left(\frac{x}{4} \right)}

    2. Integrate term-by-term:

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

        ∫(−sin⁡(x4)cos⁡2(x4)) dx=−∫sin⁡(x4)cos⁡2(x4) dx\int \left(- \sin{\left(\frac{x}{4} \right)} \cos^{2}{\left(\frac{x}{4} \right)}\right)\, dx = - \int \sin{\left(\frac{x}{4} \right)} \cos^{2}{\left(\frac{x}{4} \right)}\, dx

        1. Let u=cos⁡(x4)u = \cos{\left(\frac{x}{4} \right)}.

          Then let du=−sin⁡(x4)dx4du = - \frac{\sin{\left(\frac{x}{4} \right)} dx}{4} and substitute −4du- 4 du:

          ∫(−4u2) du\int \left(- 4 u^{2}\right)\, du

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

            ∫u2 du=−4∫u2 du\int u^{2}\, du = - 4 \int u^{2}\, du

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

              ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

            So, the result is: −4u33- \frac{4 u^{3}}{3}

          Now substitute uu back in:

          −4cos⁡3(x4)3- \frac{4 \cos^{3}{\left(\frac{x}{4} \right)}}{3}

        So, the result is: 4cos⁡3(x4)3\frac{4 \cos^{3}{\left(\frac{x}{4} \right)}}{3}

      1. Let u=x4u = \frac{x}{4}.

        Then let du=dx4du = \frac{dx}{4} and substitute 4du4 du:

        ∫4sin⁡(u) du\int 4 \sin{\left(u \right)}\, du

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

          ∫sin⁡(u) du=4∫sin⁡(u) du\int \sin{\left(u \right)}\, du = 4 \int \sin{\left(u \right)}\, du

          1. The integral of sine is negative cosine:

            ∫sin⁡(u) du=−cos⁡(u)\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}

          So, the result is: −4cos⁡(u)- 4 \cos{\left(u \right)}

        Now substitute uu back in:

        −4cos⁡(x4)- 4 \cos{\left(\frac{x}{4} \right)}

      The result is: 4cos⁡3(x4)3−4cos⁡(x4)\frac{4 \cos^{3}{\left(\frac{x}{4} \right)}}{3} - 4 \cos{\left(\frac{x}{4} \right)}

    Method #3

    1. Rewrite the integrand:

      (1−cos⁡2(x4))sin⁡(x4)=−sin⁡(x4)cos⁡2(x4)+sin⁡(x4)\left(1 - \cos^{2}{\left(\frac{x}{4} \right)}\right) \sin{\left(\frac{x}{4} \right)} = - \sin{\left(\frac{x}{4} \right)} \cos^{2}{\left(\frac{x}{4} \right)} + \sin{\left(\frac{x}{4} \right)}

    2. Integrate term-by-term:

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

        ∫(−sin⁡(x4)cos⁡2(x4)) dx=−∫sin⁡(x4)cos⁡2(x4) dx\int \left(- \sin{\left(\frac{x}{4} \right)} \cos^{2}{\left(\frac{x}{4} \right)}\right)\, dx = - \int \sin{\left(\frac{x}{4} \right)} \cos^{2}{\left(\frac{x}{4} \right)}\, dx

        1. Let u=cos⁡(x4)u = \cos{\left(\frac{x}{4} \right)}.

          Then let du=−sin⁡(x4)dx4du = - \frac{\sin{\left(\frac{x}{4} \right)} dx}{4} and substitute −4du- 4 du:

          ∫(−4u2) du\int \left(- 4 u^{2}\right)\, du

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

            ∫u2 du=−4∫u2 du\int u^{2}\, du = - 4 \int u^{2}\, du

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

              ∫u2 du=u33\int u^{2}\, du = \frac{u^{3}}{3}

            So, the result is: −4u33- \frac{4 u^{3}}{3}

          Now substitute uu back in:

          −4cos⁡3(x4)3- \frac{4 \cos^{3}{\left(\frac{x}{4} \right)}}{3}

        So, the result is: 4cos⁡3(x4)3\frac{4 \cos^{3}{\left(\frac{x}{4} \right)}}{3}

      1. Let u=x4u = \frac{x}{4}.

        Then let du=dx4du = \frac{dx}{4} and substitute 4du4 du:

        ∫4sin⁡(u) du\int 4 \sin{\left(u \right)}\, du

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

          ∫sin⁡(u) du=4∫sin⁡(u) du\int \sin{\left(u \right)}\, du = 4 \int \sin{\left(u \right)}\, du

          1. The integral of sine is negative cosine:

            ∫sin⁡(u) du=−cos⁡(u)\int \sin{\left(u \right)}\, du = - \cos{\left(u \right)}

          So, the result is: −4cos⁡(u)- 4 \cos{\left(u \right)}

        Now substitute uu back in:

        −4cos⁡(x4)- 4 \cos{\left(\frac{x}{4} \right)}

      The result is: 4cos⁡3(x4)3−4cos⁡(x4)\frac{4 \cos^{3}{\left(\frac{x}{4} \right)}}{3} - 4 \cos{\left(\frac{x}{4} \right)}

  3. Now simplify:

    −3cos⁡(x4)+cos⁡(3x4)3- 3 \cos{\left(\frac{x}{4} \right)} + \frac{\cos{\left(\frac{3 x}{4} \right)}}{3}

  4. Add the constant of integration:

    −3cos⁡(x4)+cos⁡(3x4)3+constant- 3 \cos{\left(\frac{x}{4} \right)} + \frac{\cos{\left(\frac{3 x}{4} \right)}}{3}+ \mathrm{constant}

The answer is:

−3cos⁡(x4)+cos⁡(3x4)3+constant- 3 \cos{\left(\frac{x}{4} \right)} + \frac{\cos{\left(\frac{3 x}{4} \right)}}{3}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                 3/x\
 |                             4*cos |-|
 |    3/x\               /x\         \4/
 | sin |-| dx = C - 4*cos|-| + ---------
 |     \4/               \4/       3    
 |                                      
/
∫sin⁡3(x4) dx=C+4cos⁡3(x4)3−4cos⁡(x4)\int \sin^{3}{\left(\frac{x}{4} \right)}\, dx = C + \frac{4 \cos^{3}{\left(\frac{x}{4} \right)}}{3} - 4 \cos{\left(\frac{x}{4} \right)}
Simplify
The answer [src] 
                    3/p\
               4*cos |-|
8        /p\         \2/
- - 4*cos|-| + ---------
3        \2/       3
4cos⁡3(p2)3−4cos⁡(p2)+83\frac{4 \cos^{3}{\left(\frac{p}{2} \right)}}{3} - 4 \cos{\left(\frac{p}{2} \right)} + \frac{8}{3} 
=
= 
                    3/p\
               4*cos |-|
8        /p\         \2/
- - 4*cos|-| + ---------
3        \2/       3
4cos⁡3(p2)3−4cos⁡(p2)+83\frac{4 \cos^{3}{\left(\frac{p}{2} \right)}}{3} - 4 \cos{\left(\frac{p}{2} \right)} + \frac{8}{3} 
8/3 - 4*cos(p/2) + 4*cos(p/2)^3/3

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

    © Mister Exam – Calculator