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x^3cosxdx
Solving integrals/ x^3cosxdx

Integral of x^3cosxdx dx

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∫01x3cos⁡(x)1 dx\int\limits_{0}^{1} x^{3} \cos{\left(x \right)} 1\, dx 
Integral(x^3*cos(x)*1, (x, 0, 1))
Detail solution

  1. Use integration by parts:

    ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

    Let u(x)=x3u{\left(x \right)} = x^{3} and let dv⁡(x)=cos⁡(x)\operatorname{dv}{\left(x \right)} = \cos{\left(x \right)}.

    Then du⁡(x)=3x2\operatorname{du}{\left(x \right)} = 3 x^{2}.

    To find v(x)v{\left(x \right)}:

    1. The integral of cosine is sine:

      ∫cos⁡(x) dx=sin⁡(x)\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}

    Now evaluate the sub-integral.

  2. Use integration by parts:

    ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

    Let u(x)=3x2u{\left(x \right)} = 3 x^{2} and let dv⁡(x)=sin⁡(x)\operatorname{dv}{\left(x \right)} = \sin{\left(x \right)}.

    Then du⁡(x)=6x\operatorname{du}{\left(x \right)} = 6 x.

    To find v(x)v{\left(x \right)}:

    1. The integral of sine is negative cosine:

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

    Now evaluate the sub-integral.

  3. Use integration by parts:

    ∫u⁡dv⁡=u⁡v⁡−∫v⁡du⁡\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}

    Let u(x)=−6xu{\left(x \right)} = - 6 x and let dv⁡(x)=cos⁡(x)\operatorname{dv}{\left(x \right)} = \cos{\left(x \right)}.

    Then du⁡(x)=−6\operatorname{du}{\left(x \right)} = -6.

    To find v(x)v{\left(x \right)}:

    1. The integral of cosine is sine:

      ∫cos⁡(x) dx=sin⁡(x)\int \cos{\left(x \right)}\, dx = \sin{\left(x \right)}

    Now evaluate the sub-integral.

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

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

    1. The integral of sine is negative cosine:

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

    So, the result is: 6cos⁡(x)6 \cos{\left(x \right)}

  5. Add the constant of integration:

    x3sin⁡(x)+3x2cos⁡(x)−6xsin⁡(x)−6cos⁡(x)+constantx^{3} \sin{\left(x \right)} + 3 x^{2} \cos{\left(x \right)} - 6 x \sin{\left(x \right)} - 6 \cos{\left(x \right)}+ \mathrm{constant}

The answer is:

x3sin⁡(x)+3x2cos⁡(x)−6xsin⁡(x)−6cos⁡(x)+constantx^{3} \sin{\left(x \right)} + 3 x^{2} \cos{\left(x \right)} - 6 x \sin{\left(x \right)} - 6 \cos{\left(x \right)}+ \mathrm{constant}

The answer (Indefinite) [src] 
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 |  3                               3                          2       
 | x *cos(x)*1 dx = C - 6*cos(x) + x *sin(x) - 6*x*sin(x) + 3*x *cos(x)
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(x3−6 x) sin⁡x+(3 x2−6) cos⁡x\left(x^3-6\,x\right)\,\sin x+\left(3\,x^2-6\right)\,\cos x
Simplify
The graph
0.001.000.100.200.300.400.500.600.700.800.905-10
Plot the graph f(x)
The answer [src] 
6 - 5*sin(1) - 3*cos(1)
−5 sin⁡1−3 cos⁡1+6-5\,\sin 1-3\,\cos 1+6 
=
= 
6 - 5*sin(1) - 3*cos(1)
−5sin⁡(1)−3cos⁡(1)+6- 5 \sin{\left(1 \right)} - 3 \cos{\left(1 \right)} + 6
Упростить
Numerical answer [src] 
0.171738158356098
0.171738158356098
The graph
Integral of x^3cosxdx dx

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

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