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Solving integrals/ x*sinx^2dx

Integral of x*sinx^2dx dx

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∫01xsin⁡2(x)1 dx\int\limits_{0}^{1} x \sin^{2}{\left(x \right)} 1\, dx 
Integral(x*sin(x)^2*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)=xu{\left(x \right)} = x and let dv⁡(x)=sin⁡2(x)\operatorname{dv}{\left(x \right)} = \sin^{2}{\left(x \right)}.

    Then du⁡(x)=1\operatorname{du}{\left(x \right)} = 1.

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

    1. Rewrite the integrand:

      sin⁡2(x)=12−cos⁡(2x)2\sin^{2}{\left(x \right)} = \frac{1}{2} - \frac{\cos{\left(2 x \right)}}{2}

    2. Integrate term-by-term:

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

        ∫12 dx=x2\int \frac{1}{2}\, dx = \frac{x}{2}

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

        ∫(−cos⁡(2x)2) dx=−∫cos⁡(2x) dx2\int \left(- \frac{\cos{\left(2 x \right)}}{2}\right)\, dx = - \frac{\int \cos{\left(2 x \right)}\, dx}{2}

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

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

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

            ∫cos⁡(u)2 du=∫cos⁡(u) du2\int \frac{\cos{\left(u \right)}}{2}\, du = \frac{\int \cos{\left(u \right)}\, du}{2}

            1. The integral of cosine is sine:

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

            So, the result is: sin⁡(u)2\frac{\sin{\left(u \right)}}{2}

          Now substitute uu back in:

          sin⁡(2x)2\frac{\sin{\left(2 x \right)}}{2}

        So, the result is: −sin⁡(2x)4- \frac{\sin{\left(2 x \right)}}{4}

      The result is: x2−sin⁡(2x)4\frac{x}{2} - \frac{\sin{\left(2 x \right)}}{4}

    Now evaluate the sub-integral.

  2. Integrate term-by-term:

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

      ∫x2 dx=∫x dx2\int \frac{x}{2}\, dx = \frac{\int x\, dx}{2}

      1. The integral of xnx^{n} is xn+1n+1\frac{x^{n + 1}}{n + 1} when n≠−1n \neq -1:

        ∫x dx=x22\int x\, dx = \frac{x^{2}}{2}

      So, the result is: x24\frac{x^{2}}{4}

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

      ∫(−sin⁡(2x)4) dx=−∫sin⁡(2x) dx4\int \left(- \frac{\sin{\left(2 x \right)}}{4}\right)\, dx = - \frac{\int \sin{\left(2 x \right)}\, dx}{4}

      1. There are multiple ways to do this integral.

        Method #1

        1. Let u=2xu = 2 x.

          Then let du=2dxdu = 2 dx and substitute du2\frac{du}{2}:

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

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

            ∫sin⁡(u)2 du=∫sin⁡(u) du2\int \frac{\sin{\left(u \right)}}{2}\, du = \frac{\int \sin{\left(u \right)}\, du}{2}

            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: −cos⁡(u)2- \frac{\cos{\left(u \right)}}{2}

          Now substitute uu back in:

          −cos⁡(2x)2- \frac{\cos{\left(2 x \right)}}{2}

        Method #2

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

          ∫2sin⁡(x)cos⁡(x) dx=2∫sin⁡(x)cos⁡(x) dx\int 2 \sin{\left(x \right)} \cos{\left(x \right)}\, dx = 2 \int \sin{\left(x \right)} \cos{\left(x \right)}\, dx

          1. There are multiple ways to do this integral.

            Method #1

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

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

              ∫u du\int u\, du

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

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

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

                  ∫u du=u22\int u\, du = \frac{u^{2}}{2}

                So, the result is: −u22- \frac{u^{2}}{2}

              Now substitute uu back in:

              −cos⁡2(x)2- \frac{\cos^{2}{\left(x \right)}}{2}

            Method #2

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

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

              ∫u du\int u\, du

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

                ∫u du=u22\int u\, du = \frac{u^{2}}{2}

              Now substitute uu back in:

              sin⁡2(x)2\frac{\sin^{2}{\left(x \right)}}{2}

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

      So, the result is: cos⁡(2x)8\frac{\cos{\left(2 x \right)}}{8}

    The result is: x24+cos⁡(2x)8\frac{x^{2}}{4} + \frac{\cos{\left(2 x \right)}}{8}

  3. Now simplify:

    x24−xsin⁡(2x)4−cos⁡(2x)8\frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} - \frac{\cos{\left(2 x \right)}}{8}

  4. Add the constant of integration:

    x24−xsin⁡(2x)4−cos⁡(2x)8+constant\frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} - \frac{\cos{\left(2 x \right)}}{8}+ \mathrm{constant}

The answer is:

x24−xsin⁡(2x)4−cos⁡(2x)8+constant\frac{x^{2}}{4} - \frac{x \sin{\left(2 x \right)}}{4} - \frac{\cos{\left(2 x \right)}}{8}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                                                     
 |                       2                              
 |      2               x    cos(2*x)     /x   sin(2*x)\
 | x*sin (x)*1 dx = C - -- - -------- + x*|- - --------|
 |                      4       8         \2      4    /
/
−2 x sin⁡(2 x)+cos⁡(2 x)−2 x28-{{2\,x\,\sin \left(2\,x\right)+\cos \left(2\,x\right)-2\,x^2 }\over{8}}
Simplify
The answer [src] 
       2                   
1   sin (1)   cos(1)*sin(1)
- + ------- - -------------
4      4            2
18−2 sin⁡2+cos⁡2−28{{1}\over{8}}-{{2\,\sin 2+\cos 2-2}\over{8}} 
=
= 
       2                   
1   sin (1)   cos(1)*sin(1)
- + ------- - -------------
4      4            2
−sin⁡(1)cos⁡(1)2+sin⁡2(1)4+14- \frac{\sin{\left(1 \right)} \cos{\left(1 \right)}}{2} + \frac{\sin^{2}{\left(1 \right)}}{4} + \frac{1}{4}
Упростить
Numerical answer [src] 
0.199693997861972
0.199693997861972

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

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