Search results

for 2, 32x-1 = eu where u = (2x-1)ln3 now du/dx = 2 ln3 so dx/du = 1/2ln3 Int 32x-1 .dx = 1/2ln3 Int eu .du = 1/2ln3 eu + C = 32x-1/2ln3 + C

yeah, I heard about that. can't wait :)

cos(x + n.pi + pi) = cos(x + n.pi)cospi + sin(x+n.pi)sin pi = -cos(x+n.pi) (-1)^(n+1) cosx = -(-1)^n cos x -cos(x+n.pi) = -(-1)^n cos x cos(x+n.pi) = (-1)^n cos x so if true for n, then true for n+1 if n = 1 cos(x+pi) = cosx cospi - sinx sinpi = -1cosx = RHS so true for all n>= 1

lol :p just do normal chain rule differentiation to each part, let u = (x+8), so du/dx = 1 you get d(b2 + u2)-1/du = -2u(b2 + u2)-2 and likewise for the other part, then add them

I haven't done enough questions to give you good general tips, or what type of questions usually come up. Just use logic to figure out when to multiply or divide by a factorial, or n, or whatever else. If you post up some questions maybe I could help more.

what's so ridiculous about it?

sinb = sin(90 - 2b + 30) = sin(180-b) b = 120 - 2b or 180 - b = 120 - 2b 3b = 120 b = 40 or 60 = -b b = -60 actually I think it's 40 +-120k, -60 +-360k

the answer is 0.8809896803 (guess and check. I have no idea how to solve this.)

can you post an example of a question that you find confusing? In general, for probability just find the probability of one case of a certain event occuring, and multiply by the number of ways it can happen.

if you haven't solved it yet d/d@ a sin(@ - y) - (x + a)sin y = 0 = a(1 - dy/d@)cos(@ - y) - dy/d@(x+a)cos y = 0 a cos(@ - y) = dy/d@ [a cos(@ - y) + (x+a)cos y] dy/d@ = a cos(@ - y)/[a cos(@ - y) + (x+a)cos y] d@/dt = u/a d@ = u dt/a a dy/u dt = a cos(@ - y)/[a cos(@ - y) + (x+a)cos y] dy/dt...

for question 2 just do implicit differentiation with respect to @ and substitute in udt/a for d@. I'll do the working if you want for question 1, I'm assuming the @ in (1/2 @2) is the d@/dt one? d(1/2 @'2)/d@' = @' d(1/2 @2) = @' d@' so L(1/2 @2)/d@ = L[d@/dt * d(d@/dt)]/d@ now ds = Ld@ so...

x = ln (1/2)

it's just the number of odd numbers between 1 and 30 (which is 30/2)

for 1., 2. and 4., simplify the RHS by expanding and factorising and it becomes pretty easy to see how to get from LHS to RHS. I find this helps in many prove/show questions for the 3rd, basic expansion using 3x = (2x + x) shouldn't take too long

Re: 回复: Re: 回复: Re: Integration of cotx if you didn't know to do it though, I would imagine it would be a pretty complex method too - basically solving a differential equation (?) for f'(x) = f(x)/cosx

Re: 回复: Re: Integration of cotx but how would you know to put (sec x + tan x) in?

sec[x] is the only hard one of those my way of doing it would be: write sec[x] as cosec[pi/4 - x] = Int 1/sin[pi/4 - x] = 1/2 Int 1/sin[pi/8 - x/2]cos[pi/8 - x/2] let u = [pi/8 - x/2] 1/2 Int (sin2u + cos2u)/(sinu cosu) .dx/du .du dx/du = 1/(d(pi/8 - x/2)/dx) = 1/(-1/2) = -2 - Int sinu/cosu +...

oh wow. how did I not see that?

ok, fair enough

I think this should be -i.sin^3(x)cos^3(x) +sin^4(x)cos^2(x) since i3 = -i, and i4 = 1