Integral of x*arcsin(2x) dx
The solution
Detail solution
-
Use integration by parts:
∫udv=uv−∫vdu
Let u(x)=asin(2x) and let dv(x)=x.
Then du(x)=1−4x22.
To find v(x):
-
The integral of xn is n+1xn+1 when n=−1:
∫xdx=2x2
Now evaluate the sub-integral.
SqrtQuadraticDenomRule(a=1, b=0, c=-4, coeffs=[1, 0, 0], context=x**2/sqrt(1 - 4*x**2), symbol=x)
-
Add the constant of integration:
2x2asin(2x)+8x1−4x2−16asin(2x)+constant
The answer is:
2x2asin(2x)+8x1−4x2−16asin(2x)+constant
The answer (Indefinite)
[src]
__________
/ 2 / 2
| asin(2*x) x *asin(2*x) x*\/ 1 - 4*x
| x*asin(2*x) dx = C - --------- + ------------ + ---------------
| 16 2 8
/
2x2arcsin(2x)−16arcsin(2x)+8x1−4x2
The graph
___
7*asin(2) I*\/ 3
--------- + -------
16 8
167arcsin2+23i
=
___
7*asin(2) I*\/ 3
--------- + -------
16 8
167asin(2)+83i
(0.687505232320913 - 0.359379330459425j)
(0.687505232320913 - 0.359379330459425j)
Use the examples entering the upper and lower limits of integration.