# Evaluate the indefinite integral integrate of (x)(arcsin(x))dx

You need to use integration by parts formula such that:

int udv = uv - int vdu

Selecting u = arcsin x  and dv = x  yields:

u = arcsin x => du = 1/(sqrt(1-x^2))

dv = xdx => v = x^2/2

int x arcsin x dx = (x^2/2)arcsin x - int x^2/(2sqrt(1 - x^2)) dx

You need to come up with the following substitution

sin t = x => cost dt= dx

int x^2/(2sqrt(1 - x^2)) dx = int sin^2 t/(2sqrt(1 - sin^2 t)) cos t dt

int sin^2 t/(2sqrt(1 - sin^2 t)) cos t dt = int sin^2 t/(2sqrt(cos^2 t)) cos t dt

int sin^2 t/(2sqrt(cos^2 t)) cos t dt = int sin^2 t/(2cos t) cos t dt

You need to use half angle formula such that:

sin^2 t = (1 - cos 2t)/2

intint sin^2 t/2 dt = int (1 - cos 2t)/4 dt

Using the linearity of integrals yields:

int (1 - cos 2t)/4 dt = int1/4 dt- int (cos 2t)/4 dt

int (1 - cos 2t)/4 dt = 1/4 t - (1/4)(sin 2t)/2 + c

Substituting back arcsin x  for t yields:

int x^2/(2sqrt(1 - x^2)) dx = (1/4)(arcsin x- (sin(2arcsin x))/2) + c

int x arcsin x dx = (x^2/2)arcsin x - (1/4)(arcsin x- (sin(2arcsin x))/2) + c

Hence, evaluating the given indefinite integral yields int x arcsin x dx = (x^2/2)arcsin x - (1/4)(arcsin x- (sin(2arcsin x))/2) + c.

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