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Solving integrals/ sin2t*cost

Integral of sin2t*cost dx

Limits of integration:

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The solution

You have entered [src] 
 pi                   
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 |  sin(2*t)*cos(t) dt
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∫0πsin⁡(2t)cos⁡(t) dt\int\limits_{0}^{\pi} \sin{\left(2 t \right)} \cos{\left(t \right)}\, dt 
Integral(sin(2*t)*cos(t), (t, 0, pi))
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⁡(t)cos⁡2(t) dt=2∫sin⁡(t)cos⁡2(t) dt\int 2 \sin{\left(t \right)} \cos^{2}{\left(t \right)}\, dt = 2 \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 \left(- u^{2}\right)\, du

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

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

    Method #2

    1. Rewrite the integrand:

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

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

      ∫2sin⁡(t)cos⁡2(t) dt=2∫sin⁡(t)cos⁡2(t) dt\int 2 \sin{\left(t \right)} \cos^{2}{\left(t \right)}\, dt = 2 \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 \left(- u^{2}\right)\, du

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

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

  2. Add the constant of integration:

    −2cos⁡3(t)3+constant- \frac{2 \cos^{3}{\left(t \right)}}{3}+ \mathrm{constant}

The answer is:

−2cos⁡3(t)3+constant- \frac{2 \cos^{3}{\left(t \right)}}{3}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                              3   
 |                          2*cos (t)
 | sin(2*t)*cos(t) dt = C - ---------
 |                              3    
/
∫sin⁡(2t)cos⁡(t) dt=C−2cos⁡3(t)3\int \sin{\left(2 t \right)} \cos{\left(t \right)}\, dt = C - \frac{2 \cos^{3}{\left(t \right)}}{3}
Simplify
The graph
0.000.250.500.751.001.251.501.752.002.252.502.753.002-2
Plot the graph f(x)
The answer [src] 
4/3
43\frac{4}{3} 
=
= 
4/3
43\frac{4}{3} 
4/3
Numerical answer [src] 
1.33333333333333
1.33333333333333

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

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