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Solving integrals/ sin^3t

Integral of sin^3t dt

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The solution

You have entered [src] 
 POST_GRBEK_SMALL_pi          
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         |          sin (t) dt
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∫0POSTGRBEKSMALLπ2sin⁡3(t) dt\int\limits_{0}^{\frac{POST_{GRBEK SMALL \pi}}{2}} \sin^{3}{\left(t \right)}\, dt 
Integral(sin(t)^3, (t, 0, POST_GRBEK_SMALL_pi/2))
Detail solution

  1. Rewrite the integrand:

    sin⁡3(t)=(1−cos⁡2(t))sin⁡(t)\sin^{3}{\left(t \right)} = \left(1 - \cos^{2}{\left(t \right)}\right) \sin{\left(t \right)}

  2. There are multiple ways to do this integral.

    Method #1

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

      Then let du=−sin⁡(t)dtdu = - \sin{\left(t \right)} dt and substitute dudu:

      ∫(u2−1) du\int \left(u^{2} - 1\right)\, du

      1. Integrate term-by-term:

        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}

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

          ∫(−1) du=−u\int \left(-1\right)\, du = - u

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

      Now substitute uu back in:

      cos⁡3(t)3−cos⁡(t)\frac{\cos^{3}{\left(t \right)}}{3} - \cos{\left(t \right)}

    Method #2

    1. Rewrite the integrand:

      (1−cos⁡2(t))sin⁡(t)=−sin⁡(t)cos⁡2(t)+sin⁡(t)\left(1 - \cos^{2}{\left(t \right)}\right) \sin{\left(t \right)} = - \sin{\left(t \right)} \cos^{2}{\left(t \right)} + \sin{\left(t \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⁡(t)cos⁡2(t)) dt=−∫sin⁡(t)cos⁡2(t) dt\int \left(- \sin{\left(t \right)} \cos^{2}{\left(t \right)}\right)\, dt = - \int \sin{\left(t \right)} \cos^{2}{\left(t \right)}\, dt

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

          Then let du=−sin⁡(t)dtdu = - \sin{\left(t \right)} dt and substitute −du- du:

          ∫u2 du\int u^{2}\, du

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

            ∫(−u2) du=−∫u2 du\int \left(- u^{2}\right)\, du = - \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: −u33- \frac{u^{3}}{3}

          Now substitute uu back in:

          −cos⁡3(t)3- \frac{\cos^{3}{\left(t \right)}}{3}

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

      1. The integral of sine is negative cosine:

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

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

    Method #3

    1. Rewrite the integrand:

      (1−cos⁡2(t))sin⁡(t)=−sin⁡(t)cos⁡2(t)+sin⁡(t)\left(1 - \cos^{2}{\left(t \right)}\right) \sin{\left(t \right)} = - \sin{\left(t \right)} \cos^{2}{\left(t \right)} + \sin{\left(t \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⁡(t)cos⁡2(t)) dt=−∫sin⁡(t)cos⁡2(t) dt\int \left(- \sin{\left(t \right)} \cos^{2}{\left(t \right)}\right)\, dt = - \int \sin{\left(t \right)} \cos^{2}{\left(t \right)}\, dt

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

          Then let du=−sin⁡(t)dtdu = - \sin{\left(t \right)} dt and substitute −du- du:

          ∫u2 du\int u^{2}\, du

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

            ∫(−u2) du=−∫u2 du\int \left(- u^{2}\right)\, du = - \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: −u33- \frac{u^{3}}{3}

          Now substitute uu back in:

          −cos⁡3(t)3- \frac{\cos^{3}{\left(t \right)}}{3}

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

      1. The integral of sine is negative cosine:

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

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

  3. Now simplify:

    (cos⁡(2t)−5)cos⁡(t)6\frac{\left(\cos{\left(2 t \right)} - 5\right) \cos{\left(t \right)}}{6}

  4. Add the constant of integration:

    (cos⁡(2t)−5)cos⁡(t)6+constant\frac{\left(\cos{\left(2 t \right)} - 5\right) \cos{\left(t \right)}}{6}+ \mathrm{constant}

The answer is:

(cos⁡(2t)−5)cos⁡(t)6+constant\frac{\left(\cos{\left(2 t \right)} - 5\right) \cos{\left(t \right)}}{6}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                 
 |                              3   
 |    3                      cos (t)
 | sin (t) dt = C - cos(t) + -------
 |                              3   
/
cos⁡3t3−cos⁡t{{\cos ^3t}\over{3}}-\cos t
Simplify
The answer [src] 
                                  3/POST_GRBEK_SMALL_pi\
                               cos |-------------------|
2      /POST_GRBEK_SMALL_pi\       \         2         /
- - cos|-------------------| + -------------------------
3      \         2         /               3
cos⁡3(POSTGRBEKSMALLπ2)3−cos⁡(POSTGRBEKSMALLπ2)+23\frac{\cos^{3}{\left(\frac{POST_{GRBEK SMALL \pi}}{2} \right)}}{3} - \cos{\left(\frac{POST_{GRBEK SMALL \pi}}{2} \right)} + \frac{2}{3} 
=
= 
                                  3/POST_GRBEK_SMALL_pi\
                               cos |-------------------|
2      /POST_GRBEK_SMALL_pi\       \         2         /
- - cos|-------------------| + -------------------------
3      \         2         /               3
cos⁡3(POSTGRBEKSMALLπ2)3−cos⁡(POSTGRBEKSMALLπ2)+23\frac{\cos^{3}{\left(\frac{POST_{GRBEK SMALL \pi}}{2} \right)}}{3} - \cos{\left(\frac{POST_{GRBEK SMALL \pi}}{2} \right)} + \frac{2}{3}
Simplify

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

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