-8/3cos^3(t) = -1/3(4-x^2)^(3/2) which can also be done just by inspection.

Which just verifies, once again, that rates and areas ( and by implication, calculus itself) are all just really about multiplying numbers. Thanks George.

No need to do substitution. The integral of x times sqt of (4-x^2) dx = -1/2 of the integral of sqt of (4-x^2) d(-x^2+4), which will be sqt of (4-x^2)^3,done.

I like it.

-8/3cos^3(t) = -1/3(4-x^2)^(3/2) which can also be done just by inspection.

Which just verifies, once again, that rates and areas ( and by implication, calculus itself) are all just really about multiplying numbers. Thanks George.

No need to do substitution. The integral of x times sqt of (4-x^2) dx = -1/2 of the integral of sqt of (4-x^2) d(-x^2+4), which will be sqt of (4-x^2)^3,done.

oops, forgot to put 1/3 in front of the answer.

@jjunxianlu I think George knew that you can do it by inspection. He was just offering this as a unique way (over-kill) of integrating.