Mister Exam

Other calculators

Solving integrals/ x*arccos(x^2)

Integral of x*arccos(x^2) dx

Limits of integration:

from to
v

The graph:

from to

Piecewise:

The solution

You have entered [src] 
  1              
  /              
 |               
 |        / 2\   
 |  x*acos\x / dx
 |               
/                
3/4
∫341xacos⁡(x2) dx\int\limits_{\frac{3}{4}}^{1} x \operatorname{acos}{\left(x^{2} \right)}\, dx 
Integral(x*acos(x^2), (x, 3/4, 1))
Detail solution

  1. There are multiple ways to do this integral.

    Method #1

    1. Let u=x2u = x^{2}.

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

      ∫acos⁡(u)2 du\int \frac{\operatorname{acos}{\left(u \right)}}{2}\, du

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

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

        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(u)=acos⁡(u)u{\left(u \right)} = \operatorname{acos}{\left(u \right)} and let dv⁡(u)=1\operatorname{dv}{\left(u \right)} = 1.

          Then du⁡(u)=−11−u2\operatorname{du}{\left(u \right)} = - \frac{1}{\sqrt{1 - u^{2}}}.

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

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

            ∫1 du=u\int 1\, du = u

          Now evaluate the sub-integral.

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

          ∫(−u1−u2) du=−∫u1−u2 du\int \left(- \frac{u}{\sqrt{1 - u^{2}}}\right)\, du = - \int \frac{u}{\sqrt{1 - u^{2}}}\, du

          1. Let u=1−u2u = 1 - u^{2}.

            Then let du=−2ududu = - 2 u du and substitute −du2- \frac{du}{2}:

            ∫(−12u) du\int \left(- \frac{1}{2 \sqrt{u}}\right)\, du

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

              ∫1u du=−∫1u du2\int \frac{1}{\sqrt{u}}\, du = - \frac{\int \frac{1}{\sqrt{u}}\, du}{2}

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

                ∫1u du=2u\int \frac{1}{\sqrt{u}}\, du = 2 \sqrt{u}

              So, the result is: −u- \sqrt{u}

            Now substitute uu back in:

            −1−u2- \sqrt{1 - u^{2}}

          So, the result is: 1−u2\sqrt{1 - u^{2}}

        So, the result is: uacos⁡(u)2−1−u22\frac{u \operatorname{acos}{\left(u \right)}}{2} - \frac{\sqrt{1 - u^{2}}}{2}

      Now substitute uu back in:

      x2acos⁡(x2)2−1−x42\frac{x^{2} \operatorname{acos}{\left(x^{2} \right)}}{2} - \frac{\sqrt{1 - x^{4}}}{2}

    Method #2

    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)=acos⁡(x2)u{\left(x \right)} = \operatorname{acos}{\left(x^{2} \right)} and let dv⁡(x)=x\operatorname{dv}{\left(x \right)} = x.

      Then du⁡(x)=−2x1−x4\operatorname{du}{\left(x \right)} = - \frac{2 x}{\sqrt{1 - x^{4}}}.

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

      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}

      Now evaluate the sub-integral.

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

      ∫(−x31−x4) dx=−∫x31−x4 dx\int \left(- \frac{x^{3}}{\sqrt{1 - x^{4}}}\right)\, dx = - \int \frac{x^{3}}{\sqrt{1 - x^{4}}}\, dx

      1. Let u=1−x4u = 1 - x^{4}.

        Then let du=−4x3dxdu = - 4 x^{3} dx and substitute −du4- \frac{du}{4}:

        ∫(−14u) du\int \left(- \frac{1}{4 \sqrt{u}}\right)\, du

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

          ∫1u du=−∫1u du4\int \frac{1}{\sqrt{u}}\, du = - \frac{\int \frac{1}{\sqrt{u}}\, du}{4}

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

            ∫1u du=2u\int \frac{1}{\sqrt{u}}\, du = 2 \sqrt{u}

          So, the result is: −u2- \frac{\sqrt{u}}{2}

        Now substitute uu back in:

        −1−x42- \frac{\sqrt{1 - x^{4}}}{2}

      So, the result is: 1−x42\frac{\sqrt{1 - x^{4}}}{2}

  2. Add the constant of integration:

    x2acos⁡(x2)2−1−x42+constant\frac{x^{2} \operatorname{acos}{\left(x^{2} \right)}}{2} - \frac{\sqrt{1 - x^{4}}}{2}+ \mathrm{constant}

The answer is:

x2acos⁡(x2)2−1−x42+constant\frac{x^{2} \operatorname{acos}{\left(x^{2} \right)}}{2} - \frac{\sqrt{1 - x^{4}}}{2}+ \mathrm{constant}

The answer (Indefinite) [src] 
  /                       ________              
 |                       /      4     2     / 2\
 |       / 2\          \/  1 - x     x *acos\x /
 | x*acos\x / dx = C - ----------- + -----------
 |                          2             2     
/
∫xacos⁡(x2) dx=C+x2acos⁡(x2)2−1−x42\int x \operatorname{acos}{\left(x^{2} \right)}\, dx = C + \frac{x^{2} \operatorname{acos}{\left(x^{2} \right)}}{2} - \frac{\sqrt{1 - x^{4}}}{2}
Simplify
The graph
0.7501.0000.7750.8000.8250.8500.8750.9000.9250.9500.9751.0-1.0
Plot the graph f(x)
The answer [src] 
                     ___
  9*acos(9/16)   5*\/ 7 
- ------------ + -------
       32           32
−9acos⁡(916)32+5732- \frac{9 \operatorname{acos}{\left(\frac{9}{16} \right)}}{32} + \frac{5 \sqrt{7}}{32} 
=
= 
                     ___
  9*acos(9/16)   5*\/ 7 
- ------------ + -------
       32           32
−9acos⁡(916)32+5732- \frac{9 \operatorname{acos}{\left(\frac{9}{16} \right)}}{32} + \frac{5 \sqrt{7}}{32} 
-9*acos(9/16)/32 + 5*sqrt(7)/32
Numerical answer [src] 
0.139632730124282
0.139632730124282

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

    © Mister Exam – Calculator