Proof of Integrals: Find Solutions Here

In summary, the conversation discussed the use of trigonometric substitution in simplifying integrals. It was explained that trigonometric substitutions are just special cases of the technique of integration by substitution and are particularly effective in simplifying integrals. Examples were given, such as using x = sin\theta for an integral of the form \int \frac{dx}{\sqrt{a-ax^2}}, to demonstrate the usefulness of trigonometric substitutions. It was also mentioned that with practice, one can develop a sense for which substitutions to use. Additionally, a proof was provided for the integral \int \frac {1} {x^2 + a^2} = \frac {1} {a} tan^{-1}(
  • #1
lazypast
85
0
I have just been shown integrals without proof and would rather understand where they came from. I'm sure these have been posted somewhere on here but don't know what to search for. If a website have all these proofs Ill gladly look at them.

[tex]
\int \frac {1} {(x^2 - a^2)^{0.5}} = ln(x + (x^2 - a^2)^{0.5})
[/tex]

and
[tex]
\int \frac {1} {x^2 + a^2} = \frac {1} {a} tan^{-1}( \frac {x} {a} )
[/tex]

thanks
 
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  • #2
1. Let [tex] x = a\sec \theta [/tex]

2. Let [tex] x = a\tan \theta [/tex]
 
  • #3
Well, try some trigonometric substitution and see where it gets you. I think you can use [tex]x = a \sec \theta[/tex] for the first integral.

Edit: oops, too late.
 
  • #4
For the first instead of sec, you might try [itex] x=a\sinh t [/itex].

Daniel.
 
  • #5
Why are trigonometry substitutions the key, do I just need to accept that?
 
  • #6
lazypast said:
Why are trigonometry substitutions the key, do I just need to accept that?
Trigonometric substitutions are just special cases of the technique of integration by substitution. The idea behind integration is that you chose a substitution that would simplify your integral to such an extent that you can perform the integral; it is equivalent to the chain rule in differentiation. It simply turns out that trigonometric substitutions are particularly good at simplifying integrals. For example if you had something of the form;

[tex]I = \int \frac{dx}{\sqrt{a-ax^2}}[/tex]

You would use the substitution [itex]x = \sin\theta[/itex] since when you substitute you would obtain something significantly simpler, can you see why?. Once you've done a few integrals, you get a feel for which substitutions to use.
 
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  • #7
number 2

[tex]
x = atan \theta , a^2 + x^2 = a^2 + a^2 tan^2 \theta
[/tex]

[tex]
common factor of a^2, a^2(1 + tan^2 \theta)
[/tex]

is there an identity for this?
 
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  • #8
Yes, there is as a matter of fact;

[tex]1+\tan^2\theta = \sec^2\theta[/tex]
 
  • #9
lazypast said:
Why are trigonometry substitutions the key, do I just need to accept that?
From [itex]sin^2 \theta + cos^2 \theta= 1[/itex] we can derive:

[tex]cos^2 \theta= 1- sin^2 \theta[/tex]
[tex]sec^2 \theta= 1+ tan^2 \theta[/tex]
[tex]tan^2 \theta= sec^2 \theta - 1[/tex]
So if you have something involving
[tex]\sqrt{1- x^2}[/tex]
[tex]\sqrt{1+ x^2}[/tex]
[tex]\sqrt{x^2- 1}[/tex]
The substitution [itex]sin \theta= x[/itex], [itex]tan \theta= x[/itex], or [itex]sec \theta = x[/itex] will put a single square inside the square root, getting rid of the square root.

As dextercioby pointed out, for hyperbolic functions, we have the identity [itex]cosh^2 x- sinh^2 x= 1[/itex] so
[tex]cosh^2 x= 1+ sinh^2 x[/tex]
[tex]sech^2 x= tanh^2- 1 x[/tex]
[tex]tanh^2 x= 1- sech^2 x[/tex]
so the hyperbolic functions can be used similarly.
 
  • #10
Always write 'dx' in your integral. It made you forget to differentiate the substitited x = ... and plug it into the integral.
 
  • #11
The previewing drives me up the wall :[


[tex]\int \frac {1} {{(x^2 - a^2)}^{0.5}} sub-x^2=a^2sec^2 \theta [/tex]

[tex] \int \frac {1} {((a^2sec^2 \theta) - a^2)^{0.5}} [/tex]

[tex] \int \frac {1} {(a^2(sec^2 \theta - 1))^{0.5}} [/tex]

[tex] \int \frac {1} {(a^2 tan^2 \theta)^{0.5}}[/tex]

I assume the square and square root cancel out, but don't know where to go after this, I am guessin integrate?
 
  • #12
Once again, your integrals without the differential form 'dx' represent nothing.
 
  • #13
You end up with [tex] \frac{1}{a}\int \frac{1}{\tan\theta} a\sec \theta \tan \theta [/tex]
 
  • #14
I cannot see how to go from mine to yours courtrigrad, I understand the 1/a but not the rest
 
  • #15
[tex] \int \frac{1}{\sqrt{a^{2}\tan^{2}\theta}}\; dx = \int \frac{1}{a\tan \theta}\; dx = \frac{1}{a} \int \frac{1}{\tan\theta} \; dx [/tex]

where [tex] dx = a\sec \theta \tan \theta [/tex]
 
  • #16
You don't need trig sub for these, there's a much easier way for say, number 2. Remember he wants a proof, not a derivation. Just differentiate the right side, and if you get the same integral it's proven.

[tex]y= arc tan (x)[/tex]
then [tex]x=tan (y)[/tex]
[tex] \frac {dx}{dy} = sec^2 y[/tex] (proven by using quotient rule, letting u= sin y and v= cos y)
[tex](sec^2 y) = 1 + tan^2 y[/tex]
[tex] \frac {dx}{dy} = 1 + tan^2 y[/tex]

Remember [tex]tan (y) = x[/tex]

[tex] \frac {dx}{dy} = 1 + x^2[/tex] then reciprocal both sides

Therefore [tex] \frac {d}{dx} arc tan (x) = \frac {1}{1+x^2}[/tex]

Now that we know that, we shall prove your integral with The Fundamental theorem of Calculus, damn i love it.

[tex]
\int \frac {1} {x^2 + a^2} dx
[/tex]

From the denominator, take out the factor of [tex]a^2[/tex]. Now it's: [tex]
\int \frac {1} {a^2 (1+\frac {x} {a})^2} dx
[/tex]

We can take out the [tex] \frac {1}{a^2}[/tex] but just take out [tex]\frac {1}{a}[/tex] and leave the other [tex]\frac {1}{a}[/tex] inside, trust me.

Now, inside the integral we have [tex]\frac {x}{a}[/tex]. Let it equal [tex]u[/tex]

[tex]u= \frac {x}{a}[/tex]
[tex]\frac {du}{dx} = \frac {1}{a}[/tex]

Sub the [tex]\frac {du}{dx}[/tex] for [tex]\frac {1}{a}[/tex] and the [tex]u[/tex] for [tex]\frac {x}{a}[/tex] in our previous integral, we get:

[tex]\frac {1}{a} \int \frac {1}{1+u^2} \frac {du}{dx} dx[/tex]

The [tex]dx[/tex] 's cancel out, so we are left with:
[tex]\frac {1}{a} \int \frac {1}{1+u^2} du[/tex]

We know already that [tex]\int \frac {1}{1+u^2} du= arc tan (u) + C[/tex], so we solve and get:

[tex]\frac {1}{a} arc tan (u) + C[/tex]

Sub back in [tex]\frac {x}{a}[/tex] for [tex]u[/tex], we finally prove that your second integral equals :

[tex]\frac {1}{a} arc tan (\frac {x}{a}) + C[/tex] as required. Hope i helped. Btw I finally got the hang of laTex :D
 
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  • #17
That couldn't have answer my question better thanks Gib, now it just need to sink in for me.
 
  • #18
No Problem mate, any more help i gtg now but ill be back tomoro.

and...

Hootenanny said:
It simply turns out that trigonometric substitutions are particularly good at simplifying integrals. For example if you had something of the form;

[tex]I = \int \frac{dx}{\sqrt{a-ax^2}}[/tex]

You would use the substitution [itex]x = \sin\theta[/itex] since when you substitute you would obtain something significantly simpler, can you see why?. Once you've done a few integrals, you get a feel for which substitutions to use.

well actually I would just take out the [itex]a^\frac {-1}{2}[/itex] to leave me with an integral I know is [itex]arcsin (x) + C[/itex]. Took what, 5 seconds? Much simpler if you remember a few integrals. O btw if anyone is wondering, what I just said can be proven in a similar fashion to what i just did a second ago, by changing [itex]arcsin (x) = y[/itex] to [itex]\sin (x) = y[/itex], differentiating, finding the recipricol. aww heck ill just do it here, but won't both with tex...

y= arc sin x
x= sin y

dx/dy= cos y (hopefully you know this ones proof at least..)

Now since we only define arc sin in the domain -pi/2 to pi/2 (otherwise with no domain there are an infinite number of solutions for each valid x).

Anyway you may recognize to domain i set to be the 1st and 4th quadrants, where cos is positive. In these quadrants, cos y = +(1-sin^2 y)^0.5.

remember sin y = x? cos y= (1-x^2)^0.5

recipricol to be dy/dx = 1/ (1-x^2)^0.5 , as required.
 
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  • #19
nice job Gib Z.
 

Related to Proof of Integrals: Find Solutions Here

1. What is the purpose of "Proof of Integrals"?

The purpose of "Proof of Integrals" is to provide a mathematical proof or justification for the methods used to solve integrals. It helps to understand the underlying principles and concepts of integration.

2. What are the different methods used to solve integrals?

There are several methods used to solve integrals, including the substitution method, integration by parts, partial fractions, and trigonometric substitution. Each method is useful for different types of integrals and can be applied depending on the given problem.

3. How do I know which method to use for a specific integral?

The method used to solve an integral depends on the form of the integrand. For example, if the integrand contains a variable raised to a power or a trigonometric function, the substitution method may be useful. If the integrand is a product of two functions, integration by parts may be used. It is essential to practice and familiarize yourself with each method to determine which one is most suitable for a given integral.

4. Can all integrals be solved analytically?

No, not all integrals can be solved analytically. Some integrals are considered to be unsolvable or do not have a closed-form solution. In such cases, numerical methods or approximations are used to find an approximate solution.

5. How can I check if my solution to an integral is correct?

You can check the correctness of your solution by differentiating it. If the derivative of the solution is equal to the original integrand, then your solution is correct. You can also use online tools or calculators to verify your answer. Additionally, double-checking your steps and calculations can also help to ensure the accuracy of your solution.

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