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Solving integrals/ sen^4xcos^5x

Integral of sen^4xcos^5x dx

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∫00sin⁡4(x)cos⁡5(x) dx\int\limits_{0}^{0} \sin^{4}{\left(x \right)} \cos^{5}{\left(x \right)}\, dx 
Integral(sin(x)^4*cos(x)^5, (x, 0, 0))
Detail solution

  1. Rewrite the integrand:

    sin⁡4(x)cos⁡5(x)=(1−sin⁡2(x))2sin⁡4(x)cos⁡(x)\sin^{4}{\left(x \right)} \cos^{5}{\left(x \right)} = \left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin^{4}{\left(x \right)} \cos{\left(x \right)}

  2. There are multiple ways to do this integral.

    Method #1

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

      Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

      ∫(u8−2u6+u4) du\int \left(u^{8} - 2 u^{6} + u^{4}\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:

          ∫u8 du=u99\int u^{8}\, du = \frac{u^{9}}{9}

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

          ∫(−2u6) du=−2∫u6 du\int \left(- 2 u^{6}\right)\, du = - 2 \int u^{6}\, du

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

            ∫u6 du=u77\int u^{6}\, du = \frac{u^{7}}{7}

          So, the result is: −2u77- \frac{2 u^{7}}{7}

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

          ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

        The result is: u99−2u77+u55\frac{u^{9}}{9} - \frac{2 u^{7}}{7} + \frac{u^{5}}{5}

      Now substitute uu back in:

      sin⁡9(x)9−2sin⁡7(x)7+sin⁡5(x)5\frac{\sin^{9}{\left(x \right)}}{9} - \frac{2 \sin^{7}{\left(x \right)}}{7} + \frac{\sin^{5}{\left(x \right)}}{5}

    Method #2

    1. Rewrite the integrand:

      (1−sin⁡2(x))2sin⁡4(x)cos⁡(x)=sin⁡8(x)cos⁡(x)−2sin⁡6(x)cos⁡(x)+sin⁡4(x)cos⁡(x)\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin^{4}{\left(x \right)} \cos{\left(x \right)} = \sin^{8}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{6}{\left(x \right)} \cos{\left(x \right)} + \sin^{4}{\left(x \right)} \cos{\left(x \right)}

    2. Integrate term-by-term:

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

        Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

        ∫u8 du\int u^{8}\, du

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

          ∫u8 du=u99\int u^{8}\, du = \frac{u^{9}}{9}

        Now substitute uu back in:

        sin⁡9(x)9\frac{\sin^{9}{\left(x \right)}}{9}

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

        ∫(−2sin⁡6(x)cos⁡(x)) dx=−2∫sin⁡6(x)cos⁡(x) dx\int \left(- 2 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{6}{\left(x \right)} \cos{\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:

          ∫u6 du\int u^{6}\, du

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

            ∫u6 du=u77\int u^{6}\, du = \frac{u^{7}}{7}

          Now substitute uu back in:

          sin⁡7(x)7\frac{\sin^{7}{\left(x \right)}}{7}

        So, the result is: −2sin⁡7(x)7- \frac{2 \sin^{7}{\left(x \right)}}{7}

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

        Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

        ∫u4 du\int u^{4}\, du

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

          ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

        Now substitute uu back in:

        sin⁡5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

      The result is: sin⁡9(x)9−2sin⁡7(x)7+sin⁡5(x)5\frac{\sin^{9}{\left(x \right)}}{9} - \frac{2 \sin^{7}{\left(x \right)}}{7} + \frac{\sin^{5}{\left(x \right)}}{5}

    Method #3

    1. Rewrite the integrand:

      (1−sin⁡2(x))2sin⁡4(x)cos⁡(x)=sin⁡8(x)cos⁡(x)−2sin⁡6(x)cos⁡(x)+sin⁡4(x)cos⁡(x)\left(1 - \sin^{2}{\left(x \right)}\right)^{2} \sin^{4}{\left(x \right)} \cos{\left(x \right)} = \sin^{8}{\left(x \right)} \cos{\left(x \right)} - 2 \sin^{6}{\left(x \right)} \cos{\left(x \right)} + \sin^{4}{\left(x \right)} \cos{\left(x \right)}

    2. Integrate term-by-term:

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

        Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

        ∫u8 du\int u^{8}\, du

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

          ∫u8 du=u99\int u^{8}\, du = \frac{u^{9}}{9}

        Now substitute uu back in:

        sin⁡9(x)9\frac{\sin^{9}{\left(x \right)}}{9}

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

        ∫(−2sin⁡6(x)cos⁡(x)) dx=−2∫sin⁡6(x)cos⁡(x) dx\int \left(- 2 \sin^{6}{\left(x \right)} \cos{\left(x \right)}\right)\, dx = - 2 \int \sin^{6}{\left(x \right)} \cos{\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:

          ∫u6 du\int u^{6}\, du

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

            ∫u6 du=u77\int u^{6}\, du = \frac{u^{7}}{7}

          Now substitute uu back in:

          sin⁡7(x)7\frac{\sin^{7}{\left(x \right)}}{7}

        So, the result is: −2sin⁡7(x)7- \frac{2 \sin^{7}{\left(x \right)}}{7}

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

        Then let du=cos⁡(x)dxdu = \cos{\left(x \right)} dx and substitute dudu:

        ∫u4 du\int u^{4}\, du

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

          ∫u4 du=u55\int u^{4}\, du = \frac{u^{5}}{5}

        Now substitute uu back in:

        sin⁡5(x)5\frac{\sin^{5}{\left(x \right)}}{5}

      The result is: sin⁡9(x)9−2sin⁡7(x)7+sin⁡5(x)5\frac{\sin^{9}{\left(x \right)}}{9} - \frac{2 \sin^{7}{\left(x \right)}}{7} + \frac{\sin^{5}{\left(x \right)}}{5}

  3. Now simplify:

    (35sin⁡4(x)−90sin⁡2(x)+63)sin⁡5(x)315\frac{\left(35 \sin^{4}{\left(x \right)} - 90 \sin^{2}{\left(x \right)} + 63\right) \sin^{5}{\left(x \right)}}{315}

  4. Add the constant of integration:

    (35sin⁡4(x)−90sin⁡2(x)+63)sin⁡5(x)315+constant\frac{\left(35 \sin^{4}{\left(x \right)} - 90 \sin^{2}{\left(x \right)} + 63\right) \sin^{5}{\left(x \right)}}{315}+ \mathrm{constant}

The answer is:

(35sin⁡4(x)−90sin⁡2(x)+63)sin⁡5(x)315+constant\frac{\left(35 \sin^{4}{\left(x \right)} - 90 \sin^{2}{\left(x \right)} + 63\right) \sin^{5}{\left(x \right)}}{315}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                      
 |                               7         5         9   
 |    4       5             2*sin (x)   sin (x)   sin (x)
 | sin (x)*cos (x) dx = C - --------- + ------- + -------
 |                              7          5         9   
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∫sin⁡4(x)cos⁡5(x) dx=C+sin⁡9(x)9−2sin⁡7(x)7+sin⁡5(x)5\int \sin^{4}{\left(x \right)} \cos^{5}{\left(x \right)}\, dx = C + \frac{\sin^{9}{\left(x \right)}}{9} - \frac{2 \sin^{7}{\left(x \right)}}{7} + \frac{\sin^{5}{\left(x \right)}}{5}
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    Use the examples entering the upper and lower limits of integration.

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