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
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 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...
\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.
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...