Trig substitution integral x^3*sqrt(4-x^2) + triangle to evaluate cosine of inverse sine at the end.

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Trig substitution integral x^3*sqrt(4-x^2) + triangle to evaluate cosine of inverse sine at the end.

When we see a square root in the integral with a variable piece squared inside, we immediately think to do a trig substitution!

In this trig sub integral, we start by making the substitution let x=2sin(theta). The motivation here is to simplify the interior of the square root using the trig identity (cos(x))^2=1-(sin(x))^2. We transform the integral in terms of theta.

Now we have a trigonometric integral (sin(x))^3*(cos(x))^2, and we get a handle on this trig integral by splitting off a factor of (sin(x))^2 and using the same pythagorean identity (sin(x))^2=1-(cos(x))^2.

This results in two integrals, each consisting of a power of cosine with a sine next to it. We can guess the antiderivatives using the chain rule backwards, and we arrive at the total antiderivative in terms of theta.

Finally, we have to transform the antiderivative in terms of x, but this requires evaluating the cosine of an inverse sine. We draw a triangle to evaluate cos(arcsin(x/2)) and express the final antiderivative in terms of simple algebraic functions of x.