How do you evaluate \int \sqrt { x ^ { 3} } \cdot d x?

1 Answer
May 13, 2018

let I=intsqrt(x^3)dx
let x=t^2(=>x^3=t^6)
=>dx/dt=2t or dx=2tdt
putting this in the integral we get
intsqrt(x^3)dx=intsqrt(t^6)*2tdt
=intt^3*2tdt=int2t^4dt
=2*t^5/5+C
substitute for t
then required integral ,I=2*(sqrtx)^5/5+C=2x^(5/2)+C
where C is constant of integration