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