Recent content by SmartyPants

Also, I was really struggling with the concept of the "dummy variable," and the idea that that alone was reason enough to allow the substitution ##\theta=x## after having started with ##\theta=\tfrac \pi 2-x##. But in proving to myself that ##\int_0^a sin(x)\,dx=\int_0^a sin(a-x)\,dx## via a...

Ahh...so it isn't generally because this is a definite integral that the substitution in question is allowable, but rather specifically because of the fact that ##\int_0^a sin(x)\,dx=\int_0^a sin(a-x)\,dx##, and in the specific substitution ##\cos x = \sin(\frac\pi 2 - x)##, ##\frac\pi 2##...

I watched a video on YouTube of someone solving this integral, and everything seems to make sense up until the point where the author “undoes” a substitution incorrectly...or so it seems, because the author arrives at the correct answer in the end. Specifically, we get to a point where the...

##\sin x \cos 2x = \frac12 (\sin 3x - \sin x)## - this is the trig manipulation I used to get to the final solution. And thank you for confirming for me that Math DF appears to be leaving something out of the IBP calculations, or at the very least skipping a number of steps that I wasn't able...

Hmm...I doctored that image to show the entire equation before posting it...weird that it's not showing you the entire equation. At any rate, you are correct - the part that got cut off is ##\frac 1 4 \int \ln(x)\sin(x)\cos(2x)dx##. Also, the picture does show what substitutions were made for...

Agreed...but even if I could show that two different antiderivatives have the same derivative, that doesn't help me understand how Math DF performed integration by parts in this particular example. IBP is simple: ##\int u~dv = uv-\int v~du##. When performing it, one must choose a ##u## and a...

Thanks for clarifying that much Mark. That makes much more sense, as I don't know that ##ln\left | sec~x \right |## (an antiderivative of ##tan~x##) can be manipulated into ##-ln\left | cos~x \right |## if the constants of integration are left out. That said, as I mentioned before, I'm not...

In calculating the integral ##\int{\ln\left(x\right)\,\sin\left(x\right)\,\cos\left(2\,x\right)}{\;\mathrm{d}x}##, I used a few online integral calculators to check my answer. According to one calculator, I got the correct antiderivative, but according to another (Math DF Integral Calculator)...

No doubt this problem is more easily solved using spherical coordinates than it is using Cartesian coordinates. I must admit I remember very little of what I learned about the Jacobian 20-something years ago, but I'm assuming you're using it here to transform from Cartesian to spherical...

Like the title and the summary suggest, I can derive the volume ##V=2\,\pi^{2}\,r^{2}\,R## for a ring torus - a doughnut-style toroid (one such that the major radius ##R## > the minor radius ##r##, and it therefore has a hole at the center) that is of circular cross-section. But I want to be...

It truly is fascinating that there are so many different ways one can go about calculating the volume of a sphere. I can think of 3 methods right off the top of my head that use calculus: solid of revolution, integration over spherical coordinates (as PeroK demonstrated above), and...

You might find it easier use vertical slices (instead of horizontal slices) and integrate along the x-axis. You're already using the "solid of revolution formula ##\int\limits_{\scriptsize a}^{\scriptsize b}{\pi\,(f\,(x))^{2}}{\;\mathrm{d}x}##...granted you initially posted it in the form...

OK, now that I've had a chance to look at your response, it appears I managed to distribute the negative sign to the ##xlnx## but not to the ##x##, and then mistakenly cancelled that ##x## with the ##x## outside the parentheses instead of adding them together. Indeed, when I account for the...

Awesome...thanks for the quick reply! I'm tied up at the moment, but I'll try to revisit it this evening. By the way, now that you've pointed out that integrating w/ respect to x can be done in only 3 integrals (I should have recognized that), it almost makes me not want to go looking for my...

**EDIT** Everything looked good in the preview, then I posted and saw that some stuff got cropped out along the right edge...give me some time and I'll fix it. Hello all, I"m trying to calculate shaded area, that is, the area bounded by the curves ##x=y^{2}-2, x=e^{y}, y=-1##, and ##y=1##...