Integrals that keep me up at night

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In summary: Or, simply look for$$t^4 +1 = (t^2 + at +1)(t^2 + bt +... + 1)$$and solve for a, b.In summary, the conversation focuses on solving a challenging integral involving the functions tangent and cotangent. The individual attempted to apply integration by parts and substitution, but was unsuccessful. After discussing with another individual, they reached the solution by factoring the quartic expression and using complex numbers.
  • #1
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Homework Statement
Need help finding the following indefinite integrals
Relevant Equations
-
Been struggling with a few integrals, I might post a few more once I progress further in my assignment.

$$1. \int \sqrt{tanx} + \sqrt{cotx} (dx)$$

Attempt1:
for integral 1, I try to apply integration by parts on both ##\sqrt{tanx}## and ##\sqrt{cotx}## separately, I then get
$$\int \underbrace{\sqrt{tanx}}_{\textrm{u}}\underbrace{(dx)}_{\textrm{dv}} + \int \underbrace{\sqrt{cotx}}_{\textrm{u}}\underbrace{(dx)}_{\textrm{dv}}$$

this gives,
$$=\left( x\sqrt{tanx}-\int \underbrace{x}_{\textrm{u}}.\underbrace{\frac{sec^2x}{2\sqrt{tanx}}}_{\textrm{dv}}(dx)\right) + \left(x\sqrt{cotx}-\int \underbrace{x}_{\textrm{u}}.\underbrace{\frac{(-coesc^2x)}{2\sqrt{cotx}}}_{\textrm{dv}}(dx)\right)$$

after applying integration by parts again I sadly get, RHS = LHS

$$=\left(xtanx - \left(x\sqrt{tanx}-\int \sqrt{tanx}(dx)\right)\right) + \left(x\sqrt{cotx}-\left(x\sqrt{cotx}-\int\sqrt{cotx}(dx)\right)\right)$$
$$=\int\sqrt{tanx}(dx)+\int\sqrt{cotx}(dx)$$
this basically lead me back to the beginning:(

Attempt2:
I also tried using a the substitution, ##t^2 = tan(x)## but even this brought me back to where I started,
$$\int \sqrt{tanx}+\frac{1}{\sqrt{tanx}} (dx) = \int\frac{tanx+1}{\sqrt{tanx}}(dx)$$
using the above sub,
$$\int\frac{t^2+1}{t}\frac{2t}{1+t^4}(dt) = 2\int\frac{t^2+1}{t^4+1}dt$$
now ##u=t^2##
$$=\int\frac{u+1}{u^2+1}\frac{du}{\sqrt{u}}=\int\frac{\sqrt{u}}{u^2+1}+\frac{1}{\sqrt{u}(u^2+1)}(du)$$
and finally if you use the trig sub, ##u = tan\theta## you end up with,
$$=\int \sqrt{tan\theta}+\sqrt{cot\theta}(d\theta)$$
I usually don't like asking for help while solving integrals, there is a different satisfaction one gets when they finally get the "aha" moment and then solve the problem, but these are literally getting in the way of my life, I don't want entire solutions, a hint or a reassurance that I am thinking in the right direction will do :)

I need help in these two as well, but I'll show my working for them once I get this pesky one out of the way.

$$2. \int \frac{1}{x^{1/2} + x^{1/3}} (dx)$$
$$3. \int \frac{cos(2x) - sin(2\phi)}{cos(x) - sin(\phi)} (dx)$$
 
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  • #2
Physics Slayer said:
Attempt2:
I also tried using a the substitution, ##t^2 = tan(x)## but even this brought me back to where I started,
$$\int\frac{t^2+1}{t}\frac{2t}{1+t^4}(dt) = 2\int\frac{t^2+1}{t^4+1}dt$$
This looks like progress.
Physics Slayer said:
and finally if you use the trig sub, ##u = tan\theta##
That takes you back to square one - you've undone your first substitution.
 
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  • #3
Instead:$$\int\frac{t^2+1}{t^4+1}dt = \int\frac{t^2}{t^4+1}dt + \int\frac{1}{t^4+1}dt $$
 
  • #4
PeroK said:
Instead:$$\int\frac{t^2+1}{t^4+1}dt = \int\frac{t^2}{t^4+1}dt + \int\frac{1}{t^4+1}dt $$
On second thoughts, perhaps that's not such a good idea.
 
  • #5
PeroK said:
On second thoughts, perhaps that's not such a good idea.
I thought about this, but the only think I could think of was the substitution ##u=t^2## so I could get rid of ##t^4## in the denominators, but the integral was still challenging to solve.
 
  • #6
Physics Slayer said:
I thought about this, but the only think I could think of was the substitution ##u=t^2## so I could get rid of ##t^4## in the denominators, but the integral was still challenging to solve.
$$t^4 +1 = (t^2 + \sqrt 2 t +1)(t^2 - \sqrt 2 t +1)$$
 
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  • #7
PeroK said:
$$t^4 +1 = (t^2 + \sqrt 2 t +1)(t^2 - \sqrt 2 t +1)$$
$$2\int\frac{t^2+1}{t^4+1}(dt) = \int\frac{2t^2+2}{(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)}(dt)$$
$$\int\frac{(t^2+\sqrt{t}+1)+(t^2-\sqrt{2}t+1)}{(t^2+\sqrt{2}t+1)(t^2-\sqrt{2}t+1)}(dt)=\int\frac{1}{t^2+\sqrt{2}t+1}+\frac{1}{t^2-2\sqrt{t}+1}(dt)$$
$$\int \frac{1}{(t + \frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2} + \frac{1}{(t - \frac{1}{\sqrt{2}})^2 + (\frac{1}{\sqrt{2}})^2} (dt)$$
we can use,
$$\int\frac{1}{x^2+a^2}dx = \frac{1}{a}tan^{-1}\left(\frac{x}{a}\right)$$
finally my ans is,
$$\sqrt{2}\left[tan^{-1}(\sqrt{2tanx}+1) + tan^{-1}(\sqrt{2tanx}-1)\right] +C$$

how were you able to factor ##(t^4+1)## into ##(t^2 -\sqrt{2}t + 1)(t^2 + \sqrt{2}t + 1)## ?
It's obviously correct(I only know because I multiplied the two and verified) but what method did you use?
 
  • #8
Physics Slayer said:
how were you able to factor ##(t^4+1)## into ##(t^2 -\sqrt{2}t + 1)(t^2 + \sqrt{2}t + 1)## ?
It's obviously correct(I only know because I multiplied the two and verified) but what method did you use?
I've seen that sort of thing before. I was actually looking at complex numbers first:
$$t^4 + 1 = (t^2 + i)(t^2 - i)$$but then I remembered that you can factorise a quartic into two quadratics.
 
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  • #9
PeroK said:
I've seen that sort of thing before. I was actually looking at complex numbers first:
$$t^4 + 1 = (t^2 + i)(t^2 - i)$$but then I remembered that you can factorise a quartic into two quadratics.
You can keep going with the complex numbers:
$$t^4 + 1 = (t + \frac{1}{\sqrt 2} +\frac{i}{\sqrt 2})(t + \frac{1}{\sqrt 2} - \frac{i}{\sqrt 2})(t - \frac{1}{\sqrt 2} +\frac{i}{\sqrt 2})(t - \frac{1}{\sqrt 2} - \frac{i}{\sqrt 2})$$$$ = (t^2 +\sqrt 2 t +1)(t^2 - \sqrt 2 t +1)$$Or, simply look for$$t^4 +1 = (t^2 + at +1)(t^2 + bt + 1)$$
 
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  • #10
PeroK said:
I've seen that sort of thing before. I was actually looking at complex numbers first:
$$t^4 + 1 = (t^2 + i)(t^2 - i)$$but then I remembered that you can factorise a quartic into two quadratics.
we haven't yet studied complex numbers, so I doubt this was the method they expected us to use. Either way, thanks for your help Perok.

If anybody is able to find a soln. without factorising a quartic, do let me know:smile:
 
  • #11
Physics Slayer said:
If anybody is able to find a soln. without factorising a quartic, do let me know:smile:
It's definitely the way to go. Factorise then partial fractions.
 
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  • #12
Physics Slayer said:
If anybody is able to find a soln. without factorising a quartic, do let me know:smile:
If you can cope with the language, see:
 
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  • #13
Steve4Physics said:
If you can cope with the language, see:

I'm Indian so I don't mind the heavy accent,
I usually don't search up integrals because they directly slap you with the answer, and I don't like that.
thanks for the video tho :)
 
  • #14
Physics Slayer said:
I'm Indian so I don't mind the heavy accent,
I usually don't search up integrals because they directly slap you with the answer, and I don't like that.
thanks for the video tho :)
Being a bit pedantic, can I add that both the integral in Post #1 and the integral at the start of the video are incorrect (brackets)!

The integral should, of course, be:$$\int (\sqrt{tanx} + \sqrt{cotx} )dx$$
 
  • #15
Steve4Physics said:
Being a bit pedantic, can I add that both the integral in Post #1 and the integral at the start of the video are incorrect (brackets)!

The integral should, of course, be:$$\int (\sqrt{tanx} + \sqrt{cotx} )dx$$
As far as I am aware, the notation for a indefinite integral is,
$$\int f(x)dx$$
In #1, ##f(x) = \sqrt{tanx}+\sqrt{cotx}## so I don't really see what's wrong
 
  • #16
Physics Slayer said:
As far as I am aware, the notation for a indefinite integral is,
$$\int f(x)dx$$
Agreed.

Physics Slayer said:
In #1, ##f(x) = \sqrt{tanx}+\sqrt{cotx}## so I don't really see what's wrong
Maybe it's just the convention I was taught. But I would argue that if
##f(x) = \sqrt{tanx}+\sqrt{cotx}##
then in general
##f(x). a## should be be written as
##(\sqrt{tanx}+\sqrt{cotx}).a## and not as
##\sqrt{tanx}+\sqrt{cotx}.a##
because ##\sqrt{tanx}## would not get be multiplied by ##a##. (For example 1+2*3 = 7, not 9.)

This should apply whatever'##a##' is, even if it is the infinitessimal quantity ##dx##.
 
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  • #17
$$2. \int\frac{1}{x^{1/2}+x^{1/3}} dx $$
I tried both the substitutions, ##u^2=x## and ##u^3=x## and both lead to integrals I am unable to solve,

##u^2=x##
$$\int\frac{2u}{u+u^{2/3}}(du) = 2\int\frac{1}{1+u^{-1/3}}(du)=2\int\frac{u^{1/3}}{u^{1/3}+1}(du)$$
now I use the substitution ##u=t^3##
$$2\int\frac{t}{t+1}(3t^2)(dt) = 6\int\frac{t^3}{t+1}(dt) $$
I am unable to solve this final integral. I tried applying Integration by parts consecutively 3 times to get rid of the ##t^3##, but that didn't work either.

using the other substitution(##u^3=x##) I end up with,
$$3\int\frac{u^2}{u^{3/2}+u} (du) = 3\int\frac{u}{\sqrt{u}+1}(du)$$
which is again a integral I can't solve.
 
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  • #18
Physics Slayer said:
I need help in these two as well, but I'll show my working for them once I get this pesky one out of the way.

$$2. \int \frac{1}{x^{1/2} + x^{1/3}} (dx)$$
$$3. \int \frac{cos(2x) - sin(2\phi)}{cos(x) - sin(\phi)} (dx)$$
Please post these each in an individual thread when you are ready to tackle either one.
 
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  • #19
Why not ##x = u^6##?
 
  • #20
PeroK said:
Why not ##x = u^6##?
actually in my attempt, I first use the substitution ##x=u^2## and then the substitution ##u=t^3## which is equivalent to a single substitution ##x=t^3##

But I was since then able to solve the integral, I just had to reduce the final improper fraction into a proper fraction, it was smooth sailing from there, again thanks for you help.

$$6\int\frac{t^3}{t+1}(dt) = 6\int\frac{(t+1)(t^2-t+1)-1}{t+1}(dt)=6\int\left((t^2-t+1)-\frac{1}{t+1}\right)(dt)$$
$$=6\left(\frac{t^3}{3}-\frac{t^2}{2}+t-ln|t+1|\right)=2x^{1/2}-3x^{1/3}+6x^{1/6}-6ln|x^{1/6}+1|+C$$

SammyS said:
Please post these each in an individual thread when you are ready to tackle either one.
I thought multiple posts each with a single integral problem, would appear kind of spammy, I will most definitely encounter much harder integrals as I go ahead, I don't want to flood the forum with threads on integrals, but if that's what's supposed to be done, then I shall.
 
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  • #21
Physics Slayer said:
actually in my attempt, I first use the substitution ##x=u^2## and then the substitution ##u=t^3## which is equivalent to a single substitution ##x=t^3##

But I was since then able to solve the integral, I just had to reduce the final improper fraction into a proper fraction, it was smooth sailing from there, again thanks for you help.

$$6\int\frac{t^3}{t+1}(dt) = 6\int\frac{(t+1)(t^2-t+1)-1}{t+1}(dt)=6\int\left((t^2-t+1)-\frac{1}{t+1}\right)(dt)$$
$$=6\left(\frac{t^3}{3}-\frac{t^2}{2}+t-ln|t+1|\right)=2x^{1/2}-3x^{1/3}+6x^{1/6}-6ln|x^{1/6}+1|+C$$I thought multiple posts on just one integral would appear kind of spammy, I will most definitely encounter much harder integrals as I go ahead, I don't want to flood the forum with threads on integrals, but if that's what's supposed to be done, then I shall.
It looks to me that those two integrals I referred to are entirely different from the first one.
 
  • #22
SammyS said:
It looks to me that those two integrals I referred to are entirely different from the first one.
yes they are indeed entirely different, they are three different problems(integrals), what I was trying to say was Three threads(+) each discussing only one integral might be a little spammy, hence I thought it would be better to discuss most of them here, but if that's not allowed then I shall start a new thread.
 
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  • #23
PeroK said:
Or, simply look for$$t^4 +1 = (t^2 + at +1)(t^2 +bt + 1)$$
@Physics Slayer :
Back to problem #1.

The above quote is from Post #9 by PeroK (with a minor but important Typo fixed).

Expand the right hand side and equate corresponding coefficients.

##t^4+1 = (t^2 + at +1)(t^2 +bt + 1)##

##t^4+1=t^4+(a+b)t^3+(2+ab)t^2+(a+b)t+1##

The cubic terms (also the linear terms) give you that ##\ 0 = a+b\ .\ ## Thus ##b=-a## .

The quadratic terms give you that ##0=2+ab\ .\ ## Together with the previous you get ##a^2=2.##

Thus ##a=\pm\sqrt{2 \,} ## .
 
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Related to Integrals that keep me up at night

1. What is the purpose of integrals?

Integrals are used to find the area under a curve or the accumulation of a quantity over a given interval. They are an important tool in calculus and are used in various fields of science such as physics, engineering, and economics.

2. How do I solve integrals?

Integrals can be solved using various techniques such as substitution, integration by parts, and trigonometric identities. It is important to understand the fundamental principles of integration and practice solving different types of integrals to become proficient in solving them.

3. What is the difference between definite and indefinite integrals?

A definite integral has specific limits of integration and gives a numerical value as a result, while an indefinite integral does not have limits and gives a general solution in the form of a function. Definite integrals are used to find the area under a curve, while indefinite integrals are used to find the antiderivative of a function.

4. Can integrals be used to solve real-world problems?

Yes, integrals are used in various real-world applications such as calculating the volume of irregular shapes, determining the work done by a force, and finding the average value of a function. They are a powerful tool in solving complex problems in science and engineering.

5. Are there any common mistakes to avoid when solving integrals?

Yes, some common mistakes to avoid when solving integrals include forgetting to include the constant of integration, not simplifying the final answer, and incorrectly applying integration rules. It is important to double-check your work and practice regularly to avoid making these mistakes.

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