EDIT: let x = 5sin(u)
I thought it would be fun to see if I could find the area of a circle with a radius of 5 by taking the integral of the positive form of a circle equation and multiplying it by two (since I believe the area of the bottom half is equal to the area of the upper half, or I should be able to take a definite integral from 0 to 5 and multiply it by 4). It turns out that integrating it was not as easy as I thought it would be.
I tried u-replacement, even though it looks like nothing will cancel.
I don't think there's anything I can do with this.
It looks like the equation fits one of the inverse trigonomic integrals, but it's not a fraction.
Next, I try to integrate by parts. I can't really think of anything else.
By the way, if there's something I should have seen here that would indicate to me that integrating by parts here will not work, let me know.
I don't think it helped. I could maybe integrate by parts again, but it would probably not make it any easier to solve. What could I do to solve this?
I wouldn't consider that technical, it's just being a bit cautious, 'cause suppose you're solving the problem and by not adding the absolute value bars you could get a result which is not correct.
Besides, dunno if that mark would go, dunno how strict is the professor.