1. Verify the following identity:

2. Find the antiderivative:

I really don't understand these two, need some help

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- May 20th 2006, 03:46 PMnirva2 short calc problems
1. Verify the following identity:

2. Find the antiderivative:

I really don't understand these two, need some help - May 20th 2006, 05:49 PMThePerfectHackerQuote:

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,

- May 20th 2006, 06:02 PMThePerfectHackerQuote:

Originally Posted by**nirva**

then,

and,

Thus, by substitution,

Using the Identity,

We have,

This is a linear substitution function henceforth,

Thus,

Substituting back,

- May 20th 2006, 06:08 PMticbolQuote:

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).

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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.