# 2 short calc problems

• May 20th 2006, 03:46 PM
nirva
2 short calc problems
1. Verify the following identity:
$\displaystyle \ -ln |csc(x) + cot (x)| = \ln |csc(x) - cot(x)|$

2. Find the antiderivative:
$\displaystyle \int 6t \cos (6t^2 + 5) \cos (3t^2 - 4) dt$

I really don't understand these two, need some help
• May 20th 2006, 05:49 PM
ThePerfectHacker
Quote:

Originally Posted by nirva
1. Verify the following identity:
$\displaystyle \ -ln |csc(x) + cot (x)| = \ln |csc(x) - cot(x)|$

Note if,
$\displaystyle f'(x)=g'(x)$
Then,
$\displaystyle f(x)=g(x)+C$ for sum real number $\displaystyle C$.
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,
$\displaystyle (\ln |x|)'=1/x$
$\displaystyle (\csc x)'=-\csc x\cot x$
$\displaystyle (\cot x)'=-\csc^2x$
---------
Thus given,
$\displaystyle y=\ln| \csc x-\cot x|$
Then by the chain rule you have,
$\displaystyle y'=(\csc x-\cot x)'\cdot \frac{1}{\csc x-\cot x}$
Thus,
$\displaystyle y'=\frac{(-\csc x\cot x +\csc^2x)}{\csc x-\cot x}=\frac{\csc x(\csc x-\cot x)}{\csc x-\cot x}=\csc x$

Next given,
$\displaystyle y=-\ln|\csc x+\cot x|$
Then,
$\displaystyle y'=-(\csc x+\cot x)'\cdot \frac{1}{\csc x+\cot x}$
Thus,
$\displaystyle y'=\frac{(\csc x\cot x+\csc^2x)}{\csc x+\cot x}=\frac{\csc x(\cot x+\csc x)}{\csc x+\cot x}=\csc x$

Since their derivative match implies that the function differ by a constant. Thus,
$\displaystyle \ln |\csc x-\cot x|=-\ln|\csc x+\cot x|+C$
Let $\displaystyle x=\pi/4$, then,
$\displaystyle \csc x=\sqrt{2}$
$\displaystyle \cot x=1$
Thus, substituting these values,
$\displaystyle \ln|\sqrt{2}-1|=-\ln|1+\sqrt{2}|+C$
Note that,
$\displaystyle \sqrt{2}-1>0,1+\sqrt{2}>0$ thus,
$\displaystyle \ln (\sqrt{2}-1)=-\ln (\sqrt{2}+1)+C$
Call $\displaystyle n=\sqrt{2}-1$ then, $\displaystyle n^{-1}=\sqrt{2}+1$, then,
$\displaystyle \ln n=-\ln n^{-1}+C$
Using the rule of exponents for logarithms we have,
$\displaystyle \ln n=-(-1)\ln n+C$
Thus,
$\displaystyle \ln n=\ln n+C$
Thus, $\displaystyle C=0$
Thus,
$\displaystyle \ln |\csc x-\cot x|=-\ln|\csc x+\cot x|$

Note: The reason why if $\displaystyle n=\sqrt{2}-1$ then, $\displaystyle n^{-1}=\sqrt{2}+1$ is simple. Because,
$\displaystyle n^{-1}=\frac{1}{\sqrt{2}-1}\cdot \frac{\sqrt{2}+1}{\sqrt{2}+1}=\sqrt{2}+1$
• May 20th 2006, 06:02 PM
ThePerfectHacker
Quote:

Originally Posted by nirva
2. Find the antiderivative:
$\displaystyle \int 6t \cos (6t^2 + 5) \cos (3t^2 - 4) dt$

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

Let,
$\displaystyle x=\3t^2-4$ then,
$\displaystyle 2x+13=6t^2+5$ and,
$\displaystyle dx/dt=6t$
Thus, by substitution,
$\displaystyle \int \cos (2x+13)\cos x \frac{dx}{dt}dt=\int \cos (2x+13)\cos xdx$
Using the Identity,
$\displaystyle \cos x\cos y=(1/2)[\cos (x+y)+\cos (x-y)]$
We have,
$\displaystyle \frac{1}{2}\int \cos (3x+13) +\cos (x+13) dx$
This is a linear substitution function henceforth,
$\displaystyle \frac{1}{2}\cdot \left( \frac{1}{3}\sin (3x+13)+\sin (x+13) +C_1\right)$
Thus,
$\displaystyle \frac{1}{6} \sin (3x+13)+\frac{1}{2} \sin (x+13)+C$
Substituting back,
$\displaystyle \frac{1}{6}\sin (3t^2+1)+\frac{1}{2}\sin (t^2+9)+C$
• May 20th 2006, 06:08 PM
ticbol
Quote:

Originally Posted by nirva
1. Verify the following identity:
$\displaystyle \ -ln |csc(x) + cot (x)| = \ln |csc(x) - cot(x)|$

2. Find the antiderivative:
$\displaystyle \int 6t \cos (6t^2 + 5) \cos (3t^2 - 4) dt$

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

Here are some ways.

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.