# Math Help - Help!! Calculus Problems!

1. ## 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!

2. Do you have knowledge of complex analysis or only the basic (real) calculus?

3. 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 $\arctan$ is inverse function for $\tan$ we have,
$\arctan (\tan x)=x$ thus, the derivative (rate of change) of this is $(x)'=1$. The problem, is of course that $\pi/2$ is not in the domain of $\tan x$ thus, it is not differenciable at $x=\pi/2$ thus, I would say that the rate of change does not exist there.

4. 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, $u=\tan \frac{x}{2}$ for $-\pi
Then,
$\sin x=\frac{2u}{1+u^2}$
$\cos x=\frac{1-u^2}{1+u^2}$
$dx=\frac{2du}{1+u^2}$

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

5. 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.

6. Originally Posted by thebigshow500
Anyone know how to solve the following problems?
∫((sin x )/ x) dx Limit 0≦x≦∞
You need,
$\int^{\infty}_{0^+}\frac{\sin x}{x}dx$
Express, the sine as an infinite series and divide by $x$ to get,
$\int^{\infty}_{0^+}1-\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+...dx$
Upon, integration,
$\left x-\frac{x^3}{3\cdot 3!}+\frac{x^5}{5\cdot 5!}-\frac{x^7}{7\cdot 7!} \right|^{\infty}_{0^+}$

I just do not know what this infinite sum is equal to. It seems to be to diverge. Thus,
$\int^{\infty}_{0^+}\frac{\sin x}{x}dx$
diverges.

7. 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.

8. Strange, since for example the integral

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

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

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

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

is typically one which can be computed using complex analysis.
For the record, it does converge, to $\pi/2$
Why does it diverge according to my computer program?

10. What program are you using?

11. Originally Posted by TD!
What program are you using?
Here

This is best program out there (freeware )

12. 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).

13. 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).

I never studied complex analysis
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.

14. 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

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).

15. 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

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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