Sinpi/4 = sqrt2 / 2 . Which when squared will equal 2/4 = 1/2. Is that wrong?
Kind of Frustrated at this, finals tomorrow and I'm reviewing the last concept for the day, integrals.
Interal of sin(x)cos(x)dx with upper bound of pi/4, and lower bound 0.
I let u=sinx
and du=cos(x)dx
Which turned it into u du and then I integrated it for u^2/2. Which becomes (sin(x))^2/2 . Sin of 0 will just be 0 so I didn't worry about that bound, so I plug in pi/4 and get 1/2 / 2 = 1/4. But that's wrong. What's my mistake?
Instead of making a new thread I thought I'd just post another problem here, keeps the clutter down.
Integral of (x^2-1)/x dx
What should u be and I'm just curious on to why you know what it should be? (I'm more interested in the conceptual thinking of the problem more than the actual steps to the solution.
You pick it because when you take the derivative for du you can get an x out of it? Which is what you need to put all terms into u? But then what's the next step?
If u= 3x^2+1
du= 6x dx
You can counter balance the 6 by putting 1/6 outside the integral.
So now we have du/sqrt(u). And I guess we could change this to du u^(-1/2), and then integrate to be (1/6)(2)u^(1/2). And from there it's easy, but did I do everything up to there correctly? Or since it's 1/sqrtu, I could do ln|sqrt(u)|???
Edit again: I'm getting the wrong answer. My finished integral was 1/3*sqrt(3x^2+1) with bounds from -3 to 1. Plug in and I got 2/3 - 2/3*sqrt(7), which is wrong.