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By parts , substitution u = 1/x optionalSubstitution u = 1/x would be necessary if we want to calculate it numerically5:25 This integral can also be calculated by parts and you have already prepared parts u = sqrt(1-t^2) and dv = dt
x->1/xdx->-dx/x^2I=int[0,1](x•arcsin(x))dxt=arcsin(x)dx=cos(t)dtI=int[0,pi/2](t•sin(t)cos(t))dtI=1/2•int[0,pi/2](t•sin(2t))dtp=2tdp=2dtI=1/8•int[0,pi](p•sin(p))dpD Ip sin(p)1 -cos(p)I=1/8•(-p•cos(p))|[0,pi]+1/8•int[0,pi](cos(p))dpI=pi/8+1/8•(sin(p))|[0,pi]I=pi/8
By parts , substitution u = 1/x optional
Substitution u = 1/x would be necessary if we want to calculate it numerically
5:25 This integral can also be calculated by parts
and you have already prepared parts
u = sqrt(1-t^2) and dv = dt
x->1/x
dx->-dx/x^2
I=int[0,1](x•arcsin(x))dx
t=arcsin(x)
dx=cos(t)dt
I=int[0,pi/2](t•sin(t)cos(t))dt
I=1/2•int[0,pi/2](t•sin(2t))dt
p=2t
dp=2dt
I=1/8•int[0,pi](p•sin(p))dp
D I
p sin(p)
1 -cos(p)
I=1/8•(-p•cos(p))|[0,pi]+1/8•int[0,pi](cos(p))dp
I=pi/8+1/8•(sin(p))|[0,pi]
I=pi/8