Recent content by Archimedes II

Oh ok thanks that makes since now.

Homework Statement \displaystyle \int_{1}^{\infty} 1/(x^2+ 3 \ |sin x| +2) dx Homework Equations N/AThe Attempt at a Solution \displaystyle \int_{1}^{\infty} 1/(x^2+ 3 \ |sin x| +2) dx = \displaystyle lim_{t\rightarrow \infty} \int_{1}^{t} 1/(x^2+ 3 \ |sin x| +2) dx Side Work...

I solved it I forgot to square the sin at the end. I did't need to use a u substitution. Any regardless if you have that bound it yields a negative answer. Thanks for helping though. The answer is \pi a^4/4

Yes but the how can the bound be from a higher number to a smaller one.

Ok, I fixed it thank you very much. I didn't know how to put a bound into the intergral code.

So the code is messed up basically Integral from 0 to a [4x^2 squareroot(a^2 - x^2)]dx = 4a^4 integral from 0 to pi/2 [sin theta cos^2 theta] dtheta I'm stuck trying to change the bounds with a u substitution

Homework Statement \displaystyle \int _0 ^a 4x^2 \sqrt{a^2-x^2} dx Homework Equations \cos^2 x =1 - \sin^2 x The Attempt at a Solution Substuting Substituting x=a \sin(\theta) dx=a\cos( \theta) d\theta x=0 asin\theta=0 sin\theta=0 \theta=0 \displaystyle \int_{0}^{a} 4x^2...