# Help!! Calculus Problems!

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• Mar 25th 2006, 08:01 AM
thebigshow500
Help!! Calculus Problems!
Anyone know how to solve the following problems?

1. Integrate

∫(x-1) / (lnx) dx Limit 0≦x≦1

∫(x^Alpha)*lnx dx Limit 0≦x≦1

∫(sin x)^3 / (x^2) dx Limit 0≦x≦∞

∫((sin x )/ x) dx Limit 0≦x≦∞

2. (a) Evaluate ∫((sin x)^3) /((sin x)^3 + (cos x)^3) dx Limit 0≦x≦Pi/2

(b) What is the rate of change of f(x) = arctan (tan x) at x =Pi/2

3. A point mass m moves along the curve y = 2(X^2) due to gravity and no other external force is present. Determine the power of gravity as a function of Xo, g, t. (Xo means x knot)

Thanks!
• Mar 25th 2006, 12:40 PM
TD!
Do you have knowledge of complex analysis or only the basic (real) calculus?
• Mar 25th 2006, 04:14 PM
ThePerfectHacker
Quote:

Originally Posted by thebigshow500
Anyone know how to solve the following problems?
(b) What is the rate of change of f(x) = arctan (tan x) at x =Pi/2

Since $\displaystyle \arctan$ is inverse function for $\displaystyle \tan$ we have,
$\displaystyle \arctan (\tan x)=x$ thus, the derivative (rate of change) of this is $\displaystyle (x)'=1$. The problem, is of course that $\displaystyle \pi/2$ is not in the domain of $\displaystyle \tan x$ thus, it is not differenciable at $\displaystyle x=\pi/2$ thus, I would say that the rate of change does not exist there.
• Mar 25th 2006, 04:21 PM
ThePerfectHacker
Quote:

Originally Posted by thebigshow500
Anyone know how to solve the following problems?
2. (a) Evaluate ∫((sin x)^3) /((sin x)^3 + (cos x)^3) dx Limit 0≦x≦Pi/2

Use "Weierstrauss Substitution"
Thus, $\displaystyle u=\tan \frac{x}{2}$ for $\displaystyle -\pi<x<\pi$
Then,
$\displaystyle \sin x=\frac{2u}{1+u^2}$
$\displaystyle \cos x=\frac{1-u^2}{1+u^2}$
$\displaystyle dx=\frac{2du}{1+u^2}$

This, converts any rational function of sine and cosine into an ordinary rational function.
-----------
Let me continue,
first this function is countinous over $\displaystyle [0,\pi/2]$ thus, the integral is not improper of the second type. Thus, you may simply use the fundamental theorem of calculus.
Thus, the problem,
$\displaystyle \int^{\pi/2}_0\frac{\sin^3x}{\sin^3x+\cos^3x}dx$
transfroms after Weierstrauss substitution into,
$\displaystyle \int^1_0\frac{ \left(\frac{2u}{1+u^2} \right)^3 }{ \left( \frac{2u}{1+u^2} \right)^3+\left( \frac{1-u^2}{1+u^2} \right)^3} \cdot \frac{2}{1+u^2}du$
Multiply the top and bottom of the left fraction by $\displaystyle (1+u^2)^3$ to get,
$\displaystyle \int^1_0\frac{8u^3}{8u^3+(1-u^2)^3}\cdot \frac{2}{1+u^2}du$
Hope, this helps, it is still a mess.
• Mar 25th 2006, 04:46 PM
thebigshow500
Thank you, buddy!

There is initially a mistake on the first problem, and I have fixed it already.

I hope the other problems can be solved as well. :)
• Mar 25th 2006, 04:54 PM
ThePerfectHacker
Quote:

Originally Posted by thebigshow500
Anyone know how to solve the following problems?
∫((sin x )/ x) dx Limit 0≦x≦∞

You need,
$\displaystyle \int^{\infty}_{0^+}\frac{\sin x}{x}dx$
Express, the sine as an infinite series and divide by $\displaystyle x$ to get,
$\displaystyle \int^{\infty}_{0^+}1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+...dx$
Upon, integration,
$\displaystyle \left x-\frac{x^3}{3\cdot 3!}+\frac{x^5}{5\cdot 5!}-\frac{x^7}{7\cdot 7!} \right|^{\infty}_{0^+}$

I just do not know what this infinite sum is equal to. It seems to be to diverge. Thus,
$\displaystyle \int^{\infty}_{0^+}\frac{\sin x}{x}dx$
diverges.
• Mar 25th 2006, 10:46 PM
thebigshow500
Quote:

Originally Posted by TD!
Do you have knowledge of complex analysis or only the basic (real) calculus?

I just know the basic concepts on calculus.
For example, I know some techniques such as integration by parts, substitution, and chain rules, etc.

However, I still can't figure out how to deal with the above integral problems.
• Mar 25th 2006, 11:15 PM
TD!
Strange, since for example the integral

$\displaystyle \int^{\infty}_{0^+}\frac{\sin x}{x}dx$

is typically one which can be computed using complex analysis.
For the record, it does converge, to $\displaystyle \pi/2$
• Mar 26th 2006, 09:35 AM
ThePerfectHacker
Quote:

Originally Posted by TD!
Strange, since for example the integral

$\displaystyle \int^{\infty}_{0^+}\frac{\sin x}{x}dx$

is typically one which can be computed using complex analysis.
For the record, it does converge, to $\displaystyle \pi/2$

Why does it diverge according to my computer program?
• Mar 26th 2006, 09:38 AM
TD!
What program are you using?
• Mar 26th 2006, 09:42 AM
ThePerfectHacker
Quote:

Originally Posted by TD!
What program are you using?

Here

This is best program out there (freeware :))
• Mar 26th 2006, 09:44 AM
TD!
Apparently not 'the best' since it's saying that a convergent integral diverges ;)

I can show you the math if you like, but it'll involve complex analysis (residue calculation and complex contour integration).
• Mar 26th 2006, 09:50 AM
ThePerfectHacker
Quote:

Originally Posted by TD!
Apparently not 'the best' since it's saying that a convergent integral diverges ;)

I can show you the math if you like, but it'll involve complex analysis (residue calculation and complex contour integration).

:D

I never studied complex analysis :o
Is complex analysis really so "powerful"?

You seem to be obsessed with it, whenever anyone asks a question on these forums you always throw a complex analytic solution to his problem.
• Mar 26th 2006, 09:55 AM
TD!
You're probably referring to some integrals which happened to be doable using complex analysis. Since calculus/analysis is one of my favourite fields in math, I replied there :p

It is indeed very powerful, many 'real integrals' which are very hard or even impossible using normal calculus can be computed with complex analysis. Like this one, since sin(x)/x doesn't have a primitive function (at least not in terms of the elementary functions).
• Mar 26th 2006, 09:58 AM
ThePerfectHacker
Quote:

Originally Posted by TD!
You're probably referring to some integrals which happened to be doable using complex analysis. Since calculus/analysis is one of my favourite fields in math, I replied there :p

It is indeed very powerful, many 'real integrals' which are very hard or even impossible using normal calculus can be computed with complex analysis. Like this one, since sin(x)/x doesn't have a primitive function (at least not in terms of the elementary functions).

What did you think of expressing sin(x)/x as an infinite series and then very easy to integrate?
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