Proof of Integrals: Find Solutions Here

[lazypast]

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.

$$\int \frac {1} {(x^2 - a^2)^{0.5}} = ln(x + (x^2 - a^2)^{0.5})$$

and
$$\int \frac {1} {x^2 + a^2} = \frac {1} {a} tan^{-1}( \frac {x} {a} )$$

thanks

 

[courtrigrad]

1. Let $$x = a\sec \theta$$

2. Let $$x = a\tan \theta$$

 

[radou]

Well, try some trigonometric substitution and see where it gets you. I think you can use $$x = a \sec \theta$$ for the first integral.

Edit: oops, too late.

 

[dextercioby]

For the first instead of sec, you might try $x=a\sinh t$.

Daniel.

 

[lazypast]

Why are trigonometry substitutions the key, do I just need to accept that?

 

[Hootenanny]

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;

$$I = \int \frac{dx}{\sqrt{a-ax^2}}$$

You would use the substitution $x = \sin\theta$ 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.

 

[lazypast]

number 2

$$x = atan \theta , a^2 + x^2 = a^2 + a^2 tan^2 \theta$$

$$common factor of a^2, a^2(1 + tan^2 \theta)$$

is there an identity for this?

 

[Hootenanny]

Yes, there is as a matter of fact;

$$1+\tan^2\theta = \sec^2\theta$$

 

[HallsofIvy]

Why are trigonometry substitutions the key, do I just need to accept that?

From $sin^2 \theta + cos^2 \theta= 1$ we can derive:

$$cos^2 \theta= 1- sin^2 \theta$$
$$sec^2 \theta= 1+ tan^2 \theta$$
$$tan^2 \theta= sec^2 \theta - 1$$
So if you have something involving
$$\sqrt{1- x^2}$$
$$\sqrt{1+ x^2}$$
$$\sqrt{x^2- 1}$$
The substitution $sin \theta= x$, $tan \theta= x$, or $sec \theta = x$ 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 $cosh^2 x- sinh^2 x= 1$ so
$$cosh^2 x= 1+ sinh^2 x$$
$$sech^2 x= tanh^2- 1 x$$
$$tanh^2 x= 1- sech^2 x$$
so the hyperbolic functions can be used similarly.

 

[radou]

Always write 'dx' in your integral. It made you forget to differentiate the substitited x = ... and plug it into the integral.

 

[lazypast]

The previewing drives me up the wall :[


$$\int \frac {1} {{(x^2 - a^2)}^{0.5}} sub-x^2=a^2sec^2 \theta$$

$$\int \frac {1} {((a^2sec^2 \theta) - a^2)^{0.5}}$$

$$\int \frac {1} {(a^2(sec^2 \theta - 1))^{0.5}}$$

$$\int \frac {1} {(a^2 tan^2 \theta)^{0.5}}$$

I assume the square and square root cancel out, but don't know where to go after this, I am guessin integrate?

 

[radou]

Once again, your integrals without the differential form 'dx' represent nothing.

 

[courtrigrad]

You end up with $$\frac{1}{a}\int \frac{1}{\tan\theta} a\sec \theta \tan \theta$$

 

[lazypast]

I cannot see how to go from mine to yours courtrigrad, I understand the 1/a but not the rest

 

[courtrigrad]

$$\int \frac{1}{\sqrt{a^{2}\tan^{2}\theta}}\; dx = \int \frac{1}{a\tan \theta}\; dx = \frac{1}{a} \int \frac{1}{\tan\theta} \; dx$$

where $$dx = a\sec \theta \tan \theta$$

 

[Gib Z]

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.

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

Remember $$tan (y) = x$$

$$\frac {dx}{dy} = 1 + x^2$$ then reciprocal both sides

Therefore $$\frac {d}{dx} arc tan (x) = \frac {1}{1+x^2}$$

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

$$\int \frac {1} {x^2 + a^2} dx$$

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

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

Now, inside the integral we have $$\frac {x}{a}$$. Let it equal $$u$$

$$u= \frac {x}{a}$$
$$\frac {du}{dx} = \frac {1}{a}$$

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

$$\frac {1}{a} \int \frac {1}{1+u^2} \frac {du}{dx} dx$$

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

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

$$\frac {1}{a} arc tan (u) + C$$

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

$$\frac {1}{a} arc tan (\frac {x}{a}) + C$$ as required. Hope i helped. Btw I finally got the hang of laTex :D

 

[lazypast]

That couldn't have answer my question better thanks Gib, now it just need to sink in for me.

 

[Gib Z]

No Problem mate, any more help i gtg now but ill be back tomoro.

and...

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

$$I = \int \frac{dx}{\sqrt{a-ax^2}}$$

You would use the substitution $x = \sin\theta$ 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 $a^\frac {-1}{2}$ to leave me with an integral I know is $arcsin (x) + C$. 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 $arcsin (x) = y$ to $\sin (x) = y$, 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.

 

[courtrigrad]

nice job Gib Z.