d/dx integral from o to x of function cos(2*pi*x) du is first i do the integral and i find the derivative right. by the fundamental theorem of calculus, if there is an integral from o to x, don't i just plug the x in the function.
integral from 0 to 2pi of isin(t)e^(it)dt. I know my answer should be -pi. **I pull i out because it is a constant. My work: let u=e^(it) du=ie^(it)dt dv=sin(t) v=-cos(t) i integral sin(t)e^(it)dt= -e^(it)cos(t)+i*integral
That's the same as the integral of sin^2 x dx. Use integration by parts. Let sin x = u and sin x dx = dv v = -cos x du = cos x dx The integral is u v - integral of v du = -sinx cosx + integral of cos^2 dx which can be rewritten
how do you start this problem: integral of xe^(-2x) There are two ways: 1) Integration by parts. 2) Differentiation w.r.t. a suitably chosen parameter. Lets do 1) first. This is the "standard method", but it is often more tedious
Hello, I have some calculus homework that I can't seem to get started..at least not on the right track? I have 3 questions 1. integral of [(p^5)*(lnp)dp] I'm using the uv-integral v du formula So first, I'm finding u and I think
I have two questions, because I'm preparing for a math test on monday. 1. Use the fundamental theorem of calculus to find the derivative: (d/dt) the integral over [0, cos t] of (3/5-(u^2))du I have a feeling I will be able to find
A. Find the integral of the following function. Integral of (x√(x+1)) dx. B. Set up and evaluate the integral of (2√x) for the area of the surface generated by revolving the curve about the x-axis from 4 to 9. For part
I need to find the integral of (sin x)/ cos^3 x I let u= cos x, then got -du= sin x (Is this right correct?) I then rewrote the integral as the integral of -du/ u^3 and then rewrote that as the integral of - du(u^-3). For this
I need help with this integral. w= the integral from 0 to 5 24e^-6t cos(2t) dt. i found the the integration in the integral table. (e^ax/a^2 + b^2) (a cos bx + b sin bx) im having trouble finishing the problem from here.
Evaulate: integral 3x (sinx/cos^4x) dx I think it's sec3 x , but that from using a piece of software, so you'll have to verify that. Using uppercase 's' for the integral sign we have S 3sin(x)/cos4dx or S cos-4(x)*3sin(x)dx If you