1. Verify the following identity:
2. Find the antiderivative:
I really don't understand these two, need some help
Note if,Originally Posted by nirva
Then,
for sum real number .
This is the method which I presume your teacher wants you to prove this.
I am going to find the derivatives of these two function. Notice that,
---------
Thus given,
Then by the chain rule you have,
Thus,
Next given,
Then,
Thus,
Since their derivative match implies that the function differ by a constant. Thus,
Let , then,
Thus, substituting these values,
Note that,
thus,
Call then, , then,
Using the rule of exponents for logarithms we have,
Thus,
Thus,
Thus,
Note: The reason why if then, is simple. Because,
Here are some ways.Originally Posted by nirva
A) Prove:
-ln|cscX +cotX| = ln|cscX -cotX|
Reminders:
--log(a/b) = log(a) -log(b)
--log(1/b) = log(1) -log(b) = 0 -log(b) = -log(b)
So, -ln|cscX +cotX| == ln|1/(cscX +cotX)| ------**
Trig Identity:
sin^2(X) +cos^2(X) = 1 -------------**
Divide both sides by sin^2(X),
1 +cot^2(X) = csc^2(X)
Rearrange,
1 = csc^2(X) -cot^2(X)
Factor the RHS, [a^2 -b^2 == (a+b)(a-b)],
1 = (cscX +cotX)(cscX -cotX)
Divide both sides by (cscX +cotX),
1/(cscX +cotX) = cscX -cotX
Take the natural logs of the absolute values of both sides,
ln|1/(cscX +cotX)| = ln|cscX -cotX|
Simplifying,
-ln|cscX +cotX| = ln|cscX -cotX|
Therefore, proven.
==================================
2) Find Antiderivative of
6t*cos(6t^2 +5)cos(3t^2 -4)dt --------(i)
Here, if we can separate the two cosines, we are in good shape.
Trig Identity:
cosAcosB = (1/2)[cos(A-B) +cos(A+B)] ------**
So, in (i), if A = (6t^2 +5) and B = (3t^2 -4),
cos(6t^2 +5)cos(3t^2 -4)
= (1/2)[cos(3t^2 +9) +cos(9t^2 +1)]
And (i) becomes,
(1/2)[cos(3t^2 +9) +cos(9t^2 +1)](6t dt) ----(ia)
Now it is easy to find the antiderivative of that (ia).
INT.{(1/2)[cos(3t^2 +9) +cos(9t^2 +1)](6t dt)}
= (1/2)INT.[cos(3t^2 +9) +cos(9t^2 +1)](6t dt) ----(ii)
Let u = 3t^2 +9, so du = 6t dt
And V = 9t^2 +1, so dV = 18t dt
For simplicity, let us expand the (ii).
(You do not have to do this if you like. This is just for easy explanation.)
= (1/2)INT.[cos(3t^2 +9)](6t dt) +(1/2)INT.[cos(9t^2 +1)](6t dt)
Then, easy sailing,
= (1/2)sin(3t^2 +5) +(1/2)INT.[cos(9t^2 +1)](18t)/3 dt
= (1/2)sin(3t^2 +5) +(1/3)(1/2)INT.[cos(9t^2 +1)](18t dt)
= (1/2)sin(3t^2 +5) +(1/6)sin(9t^2 +1) -------------------answer.
For your exercise, try finding the derivative of that answer. It should match (i).
-------------------
For your other postings, if you can post only one or two problems per posting, I may try to solve them, if I have time.