Challenging Integrals in Calculus 1-2: Expand Your Problem-Solving Skills!

[dextercioby]

$$\int^2_0 \sin{x^2}\,dx$$

There's nothing hard about it and nothing interesting either, as long you've read about special functions like http://mathworld.wolfram.com/FresnelIntegrals.html

 

[dextercioby]

Compute:

$$\int_{0}^{1}\int_{0}^{1} \frac{y^2 -x^2}{(x^2 +y^2)^2} dx{}~{}dy $$

 

[phion]

There's nothing hard about it and nothing interesting either, as long you've read about special functions like http://mathworld.wolfram.com/FresnelIntegrals.html

Solve it then? Approximation techniques don't count either.

 

[h6ss]

I'm curious to see what technique you guys would use to solve $\int_{-1}^{1} \frac{\cos(x)}{\exp(1/x) + 1}dx$

 

[quantumtimeleap]

$$\int e^{-x^2} dx$$

how can this even be integrated?

You don't. You use a power series for that one. Hahaha.

 

[Calculus Master]

sin(2x)cos(2x)dx

$$\int{\sin{2x}\cos{2x}dx} = \frac{1}{2} \int{\sin{4x}} = -\frac{1}{8} \cos{4x}$$

 

[Calculus Master]

I'm just out of Calc 1, so I'm not sure if I even have the knowledge to solve this... but here's where I am now. I can't tell if I complicated it even more, or if I'm closer to getting the solution.

$$\int (r-1)^{1/5}ln(r) - r^{-1} (r-1)^{1/5} dr$$

I used u-substitution (well, r-substitution), where $$r = x^5 + 1$$. After the substitution I used integration by parts, and now I'm unsure if that was even the right path. Please let me know!

And please tell us how to do $$\int_{0}^{\infty}sin(x^2)dx$$

To integrate $$\int_{0}^{\infty}sin(x^2)dx$$ we use $$\int_{0}^{\infty}sin(x^n)dx$$ =$$\Gamma{(1 + \frac{1}{n})}\sin{\frac{\pi}{2n}}$$ = $$\frac{\sqrt{\pi}}{2\sqrt{2}}$$

 

[STEMucator]

Give this one a chug:

$$\int x \sqrt{ax^2 + bx + c}$$

 

[Curious3141]

Give this one a chug:

$$\int x \sqrt{ax^2 + bx + c}$$

I would rewrite it as $\displaystyle \frac{1}{2a}[\int(2ax + b)\sqrt{ax^2 + bx + c}dx - b\int\sqrt{ax^2 + bx + c}dx]$

The first integral is of the form $\displaystyle \int f'(x)g(f(x))dx$ and can be solved easily by either inspection or applying the sub$\displaystyle ax^2 + bx + c = u$.

The second integral is evaluated by completing the square and applying a hyperbolic sine sub.

 

[hEMU]

Disclaimer: A friend told me about this and I do not know the answer yet.
Disclaimer 2: This is not really an integration problem but very close. (If you think it should not be here, please tell me, I will remove it)

Try this one,
(dy/dx) + 3 = (dy/dx) + 2
Apparently, it does have a solution (or so I am told and yes, I am still working on the problem)

 

[zoki85]

Disclaimer: A friend told me about this and I do not know the answer yet.
Disclaimer 2: This is not really an integration problem but very close. (If you think it should not be here, please tell me, I will remove it)

Try this one,
(dy/dx) + 3 = (dy/dx) + 2
Apparently, it does have a solution (or so I am told and yes, I am still working on the problem)

Any vertical line (x=a) satisfies.
"∞ = ∞"

 

[acegikmoqsuwy]

$\displaystyle\int_0^\infty e^{-\sqrt[2015] x} dx = 2015!$

True or false? Prove your answer. Came up with this one myself :D

 

[dextercioby]

That's true, I think. I'm lazy to do the maths, but you can make an obvious substitution, then use the definition of the Euler Gamma function.

 

[cpman]

This one isn't too hard, but it is really cool:
$\int^{\infty}_{-\infty} \frac{1}{x^2 + 1} \, \mathrm{d}x$

 

[hEMU]

This one isn't too hard, but it is really cool:
$\int^{\infty}_{-\infty} \frac{1}{x^2 + 1} \, \mathrm{d}x$

Wouldn't it just be zero?
-You substitute x = tan u
-You get [arctan x] from (-inf) to (+inf) which is 0

 

[cpman]

Your first 2 steps are right, but it looks like you got a sign mixed up afterwards.

 

[hEMU]

Your first 2 steps are right, but it looks like you got a sign mixed up afterwards.

Are you implying that 0.5*pi - (- 0.5*pi) = pi so the answer should be pi?
If so, I will have to differ. arctan (-inf) = arctan (inf) = 0.5*pi. Remember that at tan(pi/2), (inf) and (-inf) co-occur.
You can also look at the area between y = 0 and y = arctan x. From (- inf) to 0 it is negative and from 0 to (inf), it is positive. Adding the areas you get 0.
If you still think that I am wrong, please tell me why.
Thanks

 

[cpman]

All that matters for this is $lim_{n \to -\infty}(\arctan n) = \frac{-\pi}{2}$ as can be seen from the graph of $y = \arctan x$. If you graph the curve $\frac{1}{x^2 -1}$, you'll see that the area under the left side of that curve is not negative, unlike with $y = \arctan x$. The curve in question is symmetrical about the y-axis, so you can reevaluate the question as $2 \int^{\infty}_{0} \frac{1}{x^2 + 1} \, \mathrm{d}x$ which is $\pi$. Also, you can solve it without using the symmetry of the curve:
It must first be broken into two improper integrals:
$\int^{0}_{-\infty} \frac{1}{x^2 + 1} \, \mathrm{d}x + \int^{\infty}_{0} \frac{1}{x^2 + 1} \, \mathrm{d}x$
The first improper integral becomes $\lim_{a \to -\infty}(\arctan a) |_{-\infty}^{0}$. This becomes $0 - lim_{a \to -\infty}(\arctan a) = \frac{\pi}{2}$. As for the second improper integral: $\lim_{b \to \infty}(\arctan b) |_{0}^{\infty} = \lim_{b \to \infty}(\arctan b) - 0 = \frac{\pi}{2}$

So, the whole integral together does evaluate to $\pi$ and not zero.

 

[dextercioby]

Well, the substitution should also be applied to the integration limits.

 

[acegikmoqsuwy]

Wouldn't it just be zero?
-You substitute x = tan u
-You get [arctan x] from (-inf) to (+inf) which is 0

arctan(inf) = pi/2 and arctan(-inf) = -pi/2 so its $\pi$

 

[hEMU]

arctan(inf) = pi/2 and arctan(-inf) = -pi/2 so its $\pi$

Ok I get it, thanks.(and yes i get why i was wrong)

 

[acegikmoqsuwy]

Find the volume of the region between $z=x^2+y^2-4$ and $z=4-x^2-y^2$ on the domain $x^2+y^2=4$ in rectangular and cylindrical coordinates. (Do cylindrical first, it is a lot easier.)

 

[phion]

To integrate $$\int_{0}^{\infty}sin(x^2)dx$$ we use $$\int_{0}^{\infty}sin(x^n)dx$$ =$$\Gamma{(1 + \frac{1}{n})}\sin{\frac{\pi}{2n}}$$ = $$\frac{\sqrt{\pi}}{2\sqrt{2}}$$

That's fantastic, thank you.

 

[HomogenousCow]

That's fantastic, thank you.

It's kind of cheating, no fun when you use formulas. I think it can be done without a formula.

 

[phion]

It's kind of cheating, no fun when you use formulas. I think it can be done without a formula.

Well, good luck!

 

[acegikmoqsuwy]

Find the volume of the region between $z=x^2+y^2-4$ and $z=4-x^2-y^2$ on the domain $x^2+y^2=4$ in rectangular and cylindrical coordinates. (Do cylindrical first, it is a lot easier.)

Oops the domain is $x^2+y^2\le 4$

 

[Mariuszek]

Inverse trig substitution not always is a good substitution for integrals with square root of quadratic
f.e
$$\int{\frac{\mbox{d}x}{x\sqrt{2x^2-2x+1}}}\\\int{\frac{\mbox{d}x}{x\sqrt{2x^2-2x-1}}}\\\int{\frac{\mbox{d}x}{x^2\left(4x^2-3\right)^2\sqrt{x^2-1}}}$$

If you want to get u substitution for ths integrals draw a curve $$y^2=ax^2+bx+c$$
and cut it with secant line
If secant line intersects curve at $$\left(\lambda,0\right)$$ or $$\left(\mu,0\right)$$ you will get third substitution
If secant line intersects curve at $$\left(0,\sqrt{c}\right)$$ or $$\left(0,-\sqrt{c}\right)$$ you will get second substitution

Assume that $$a>0$$ , draw asymptote and cut the curve with lines parallel to this asymptote
This lines will intersect point $$\left(x_{0},y_{0}\right)$$ at infinity and point $$\left(x,y\right)$$ which coordinates are rational functions of new variable
You will get first substitution in this way

You can also draw right triangle and label its sides as in inverse trig substitution
Bisect one of acute angle to get new right triangle
Calculate length of missing side using angle bisector theorem
Your u substitution is tangent of acute angle in this new right triangle

 

[Rishi Jain]

∫ sec^3 θ dx

 

[Ananya0107]

Try $$\int{\frac{(1+x^{2})dx}{(1-x^{2})\sqrt{1+x^{4}}}}$$
(forgot to put the integral sign in, it is now fixed)

How about dividing both sides by x^2 then put x-1/x = t, 1+ 1/x^2 dx = dt

 

[AdityaDev]

This one is really hard and you need to be good in calculus.
find $$\int^{\pi/2}_0(sinx-cosx)lnsinxdx$$
Believe me, its really difficult.

 

[HomogenousCow]

This one is really hard and you need to be good in calculus.
find $$\int^{\pi/2}_0(sinx-cosx)lnsinxdx$$
Believe me, its really difficult.

Looks like integrating by parts will take care of it.

 

[AdityaDev]

Looks like integrating by parts will take care of it.

No. That's just 5% of the solution. Its not easy.

 

[NeOH]

No. That's just 5% of the solution. Its not easy.

you can separate it as integral(sinx*lnsinx dx) - integral (cosx lnsinx dx)

on right side use substitution sinx=u, and use partial integration to get answer.

on left side use substitution ln(sinx)=v and you can get answer easily with beta function after the integral becomes sin^2(x)cos^-1(x) dx.

so about 5min unless you are not fluent in these methods, or if the use of beta functions is not allowed.

 

[AdityaDev]

you can separate it as integral(sinx*lnsinx dx) - integral (cosx lnsinx dx)

on right side use substitution sinx=u, and use partial integration to get answer.

on left side use substitution ln(sinx)=v and you can get answer easily with beta function after the integral becomes sin^2(x)cos^-1(x) dx.

so about 5min unless you are not fluent in these methods, or if the use of beta functions is not allowed.

I knew you would come up with that method. You are wrong.
REASON: limits.
Your answer will be filled with -infinity terms.
You cannot get the answer because the function gives infinity in lower limit. If it was an indefinite integration, your answer would have been right.
I said its a difficult question. Think again.

 

[NeOH]

I knew you would come up with that method. You are wrong.
REASON: limits.
Your answer will be filled with -infinity terms.
You cannot get the answer because the function gives infinity in lower limit. If it was an indefinite integration, your answer would have been right.
I said its a difficult question. Think again.

how? on the left integral, from the beta function you will get gamma functions and no limit problem there.

on the right side there is no limit problem either because it will become integral (ln (u)) which will integrate to u*ln(u)-u, and replacing u=sinx and it's trivial.