tan<sup>n-2</sup>x.sec<sup>2</sup>xdx = tan<sup>n-1</sup>x/(n-1)
The derivative of tan x is sec<sup>2</sup>x, so if you use substitution of u= tan x, u get du = sec<sup>2</sup>xdx
Then you are just integrating a power function of un-2
Hence, you get that result.
It's the identical result for...
The same relation will still apply.
You pull out a [cosec(x)]^2, so you'll be left with [cosec(x)]^(n-2) * d (cot(x))
Then integration by parts will solve all :)
Rearrange the equation so y's are all on one side.
x3 = -y3 + 3xy
Then divide everything by x3
So:
1 = (-y3 + 3xy)/x3
= -(y/x)3 + 3y/x2
Then take the limits of x going to infinity.
The 2nd term will become insignificant, so -(y/x)3 will tend to 1 as x tends to infinity.
Hence, y = -x
You find you horizontal tangents and your vertical tangents. Then you mark those points in.
Then, find your x and y intercepts, which should just be (0,0)
Then, you find out what happens when x -> negative infinite.
And the same for when x -> positive infinite.
once you have these points...
For part a, compare the areas under the graph. The upper rectangle (1/n) is greater than the actual area (integrate), and this area is greater than the lower rectangle (1/(n+1))
For part b, use the inequalities from part a.
Using just the inequality on the LHS:
We have 1/(n+1) < ln (1 + 1/n)...
Sorry, I'm not capable of doing individual tutoring for any uni courses yet. Maybe wait one or two years :P.
I'm assuming you are doing MATH1151 this semester.
I take a support class for this course on Tuesdays 10-11 every week. Feel free to come along.
The other support classes, I think are on...