You need to solve the integral using substitution method, such that:
`x^2 = t => 2xdx = dt => xdx = (dt)/2`
Replacing x by t, yields:
`int x*sqrt(1 - x^4) dx = (1/2)int sqrt(1 - t^2)*dt`
You need to use the next trigonometric substitution, such that:
`t = sin u => dt = cos u du`
`u = arcsin t = arcsin x^2`
`(1/2)int sqrt(1 - t^2)*dt = (1/2)*int sqrt (1 - sin^2 u)*cos u du`
You need to use the fundamental trigonometric formula `1 - sin^2 u = cos^2 u`
`(1/2)*int sqrt (1 - sin^2 u)*cos u du = (1/2)*int sqrt (cos^2 u)*cos u du`
` (1/2)*int cos u*cos u du = (1/2)*int (cos^2 u) du`
Using the half angle formula, yields:
`cos^2 u = (1 + cos 2u)/2`
`(1/2)*int (cos^2 u) du = (1/4)*int (1 + cos 2u) du`
`(1/4)*int (1 + cos 2u) du =(1/4)*intdu + (1/4)*int (cos 2u) du`
`(1/4)*int (1 + cos 2u) du =(1/4)*(u + (sin(2u))/2)`
`int x*sqrt(1 - x^4) dx = (1/4)*(arcsin (x^2) + (sin(2arcsin (x^2)))/2) + c`
Hence, evaluating the indefinite integral yields `int x*sqrt(1 - x^4) dx = (1/4)*(arcsin (x^2) + (sin(2arcsin (x^2)))/2) + c.`
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