Recent content by Chaz706

Didn't know that if you took cos(t) = u that you could allow u^2 to simply be cos^2(t) . Of course, I didn't think about that either.

The Method is substitution, but then what's u? I'm guessing the basic trig function outside the root is du. But when I derive the function within that root, I get something different. The derivative of either one ends up to be -4cos(t)sin(t) . Am I choosing the wrong U? I'm integrating...

My brain is truly fried. I know this: \int e^u du = e^u +C But what do I do if I get this: \int ae^a^u du ? Assuming a is a non-zero constant? EDIT: Never mind! I figured this out backwards. \frac {d}{du} e^a^u = ae^a^u du Thus \int ae^a^u du = e^a^u

That would have been much easier dex, I think...

Whozum: cos^2(t) - sin^2(t) = cos(2t) is the correct identity (and thanks! It may just help!) Your other identity: sin^2(t) - cos^2(t) = cos(2t) is erroneous, but not by much. sin^2(t) - cos^2(t) yields -cos(2t) This could actually help as well, however. Thank you for posting...

\int (sin(t)-cos(t)) \sqrt{cos^2(t)-sin^2(t)} dt Is there a trig idendity I can use? I've distributed that root to both terms to get: \int sin(t) \sqrt{cos^2(t)-sin^2(t)} dt - \int cos(t) \sqrt{cos^2(t)-sin^2(t)} If I take one of the terms and integrate by parts, I'm trying to put...

I've got this thanks. Thanks to a form of integration in the back of my book.

I have the limits... it's just that I can't get them right on Latex (stupid coding! I'm getting it right, it's just not displaying it that way!) Hang on... EDIT: Problem above now has working limits, and my original question as intended.

I have the following Integral \int ^1 _0 \int _0 ^\sqrt{1-x^2} \int _0 ^\sqrt{1-x^2-y^2} \frac{1}{1+(x^2)+(y^2)+(z^2)} dzdydx (With the limits working properly!) Converted to spherical Cor-ordinates, I have \int ^\frac{\pi}{2} _0 \int _0 ^\frac{\pi}{2} \int _0 ^1...

Thanks once more! :)

\int \sqrt{4-x^2}^3 dx I'm thinking integrating by parts would work, with u being that root and dv being dx, but is that the right method and direction. I've tried it and it seems more complicated.

Thanks Thanks for your help Jameson. And Older Dan too.

The General integral for a trig form works whenever the variable inside goes to the first degree. Example: Sin(x) But the general integral form for when the variable inside goes beyond the first degree doesn't work. Example: Sin(x^2), Cos(x^3) I end up getting an integral whose...