Trig. substitution can also be used if you let x=sin(θ) and dx=cos(θ)dθ. Substitute the function as the integral of (sin(θ))^3√(1-(sin(θ))^2)cos(θ)dθ=(sin(θ))^3(cos(θ))^2dθ=sin(θ)(sin(θ))^2(cos(θ))^2dθ=sin(θ)(1-(cos(θ))^2)(cos(θ))^2dθ. Using the u sub, you can let u=cos(θ) and du=-sin(θ)dθ. The u sub becomes -(1-u^2)u^2du=u^4-u^2. The integral of this is therefore, u^5/5-u^3/3+C. In order to put back in terms of x, we have to put back in terms of θ and then the x term. Therefore, (cos(θ))^5/5-(cos(θ))^3/3+C=(√(1-x^2))^5/5-(√(1-x^2))^3/3+C.
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