Recent content by mathor345

Homework Statement \int\frac{\sqrt{x^2 + 36}}{4x^2}}dx Homework Equations sqrt(a^2 + x^2) substitution for x = a tan theta The Attempt at a Solution I set x = 6 tan theta x^2 = 36 tan^2 theta dx = 6 sec^2 x \int\frac{\sqrt{36 + 36 tan \theta}}{144 tan \theta}dx \int\frac{\sqrt{36(tan...

Doh! I saw the answer as soon as I set that. Too much late night math homework after working full time really eats away at the brain, as we all see by the "backwards integration" that bigubau pointed out. Thanks guys :)

Homework Statement Evaluate. This may not require integration by parts: integral of x^5 sec(x^6) dx Homework Equations integral sec x dx = ln | sec x + tan x| + C integral u dv = uv - integral v du ... tabular integration process The Attempt at a Solution u = sec x^6 du...

Doh! Thanks guys :P

This is a question on the simplification operations. I can't for the life of me figure out how: \frac{1}{\frac{1}{2t^3}\sqrt{(\frac{1}{2t^3})^2 -1}}*(-\frac{3}{2t^4})= -\frac{3}{t\sqrt{\frac{1}{4t^6}(1 - 4t^6)}} Really, I can't figure out where (1-4t^6) is coming from! It's involved in...

How to find the limit of (e^(4/x) - 2x)^(X/2) as x--> 0+ Homework Statement \mathop{\lim}\limits_{x \to 0+} (e^{4/x} -2x)^{x/2} Homework Equations if lim ln f(x) = L then \mathop{\lim}\limits_{x \to 0+} e^{ln f(x)} = e^L The Attempt at a Solution Not too sure what my first...

Thank you.

Homework Statement differentiate y = t^{5-e} Homework Equations Power rule The Attempt at a Solution u = (5 - e) (u)t^{u - 1} = (5 - e)t^{4-e} Is this a correct usage? I'm not sure if there are any equations regarding this, but since e is a constant this should be correct right?

Thanks for the help everyone.

\int e^{3y} dy = \int e^{3x} dx \frac{e^{3x}}{3} = \frac{e^{3y}}{3} e^{3x} + C = e^{3y} + C e^{3x} + C = e^{3y} ln (e^{3x} + C) = ln (e^{3y}) ln (e^{3x} + C) = 3y \frac{1}{3} ln (e^{3x} + C) = y... I think this is the right answer, but it could very well be \frac{1}{3} ln (e^{3y} + C) =...

Homework Statement \frac{dy}{dx} = e^{3x-3y} Homework Equations \int e^{u}du = e^u + C The Attempt at a Solution \frac{dy}{dx} = \frac{e^{3x}}{e^{3y}} e^{3y} dy = e^{3x} dx ln (e^{3y}) dy = ln (e^{3x}) dx 3y dy = 3x dx Integrate... As you can see I'm doomed to get x = y

Thanks for responses, especially Mark for the detailed response. It's been a year since my last math class, so I was obviously missing some of the basics. I was completely ignoring the dx as if it had no meaning, and was a bit confused with moving constants in front of the integral sign. I...

I know, I meant that those limits are substituted for x in u (f(x)). Ok, so apparently for a definite integral \int_2^3\frac{x^5 + 1}{x^6+6x}dx must be converted into the form: \int_2^3\frac{1}{u}du So, u = x^6 + 6x so, du = 6x^5 + 6 factor out du = 6(x^5 + 1) --> \frac{1}{6}du Plugging...

Okay... I see how the 1/6 comes into the answer. I don't see why the upper and lower limit appear inside of the ln (u) though? I've never been this confused over a single problem before.

if u is x^6 + 6x then du = 6x^5 + 6 -> 6(x^5 + 1)...