1. Consider this way of differentiating...I have never seen it done...but it is an adaptation of logarithimic differentiating so I am sure other people have done it

It goes something like this...say you have $\displaystyle y=f(u(x))$....now say that $\displaystyle f(x)$ is difficult to integrate....why not then do this

[math\f^{-1}(y)=u(x)[/tex]

For example this is a way to either be a smart*** on a test if you forget or it is a way I have discovered of proving a trigonometric identity

$\displaystyle y=sin^{-1}(x)$......say you forget that derivative...well do this

$\displaystyle \sin(y)=x$...differentiating we see that

$\displaystyle y'\cdot\cos(y)=1\Rightarrow{y'=\frac{1}{cos(y)}}\R ightarrow{y'=\frac{1}{\cos(\sin^{-1}(x))}}$

The trig proof comes from the fact taht $\displaystyle y=sin^{-1}(x)\Rightarrow{y'=\frac{1}{\sqrt{1-x^2}}}$

Now since $\displaystyle y'=y'\Rightarrow{\frac{1}{\cos(\sin^{-1}(x))}=\frac{1}{\sqrt{1-x^2}}}\Rightarrow{\cos(\sin^{-1}(x))=\sqrt{1-x^2}}$

which is a famous identity...this can be done for most things

Another example would be $\displaystyle y=\ln(u(x))$

Forget how to take the derivative of ln(x)? Do this...

$\displaystyle e^{y}=u(x)\Rightarrow{y'\cdot{e^{y}}=u'(x)}\Righta rrow{y'=\frac{u'(x)}{e^{y}}}$

Now making the sub$\displaystyle y=\ln(u(x))$ we get

$\displaystyle y'=\frac{u'(x)}{e^{\ln(u(x))}}=\frac{u'(x)}{u(x)}$

which is the derivative of ln((u(x))

2. This is another messy general case integral

We have $\displaystyle \int_0^{\infty}e^{-ax}sin(bx)dx$

Ok first we realized that $\displaystyle \int_0^{\infty}f(x)=\bigg[\int{f(x)}dx\bigg]\bigg|_0^{\infty}$

So first we want to compute $\displaystyle \int{e^{-ax}\sin(bx)}dx$

So first using parts we get $\displaystyle \int{e^{-ax}\sin(bx)dx}=\frac{-\sin(bx)e^{-ax}}{a}+\frac{b}{a}\int{e^{-ax}\cos(bx)dx}$

Then using integration by parts again we get $\displaystyle \int{e^{-ax}\sin(bx)dx}=\frac{-\sin(bx)e^{-ax}}{a}+\frac{b}{a}\bigg[\frac{-\cos(bx)e^{-ax}}{a}-\frac{b}{a}\int{e^{-ax}\sin(bx)dx}\bigg]$

Distributing we get $\displaystyle \int{e^{-ax}\sin(bx)dx}=\frac{-\sin(bx)e^{-ax}}{a}-\frac{b\cos(bx)e^{-ax}}{a^2}-\frac{b^2}{a^2}\int{e^{-ax}\sin(bx)dx}$

adding the integral to each side we get $\displaystyle \frac{b^2}{a^2}\int{e^{-ax}\sin(bx)dx}+\int{e^{-ax}\sin(bx)dx}=\frac{-\sin(bx)e^{-ax}}{a}-\frac{b\cos(bx)e^{-ax}}{a^2}$

$\displaystyle \Rightarrow{(1+\frac{b^2}{a^2})\int{e^{-ax}\sin(bx)dx}=\frac{-\sin(bx)e^{-ax}}{a}-\frac{b\cos(bx)e^{-ax}}{a^2}}$

before dividing I would like to simplify the right side to get $\displaystyle e^{-ax}+\bigg[\frac{-\sin(bx)}{a}-\frac{\cos(bx)}{a^2}\bigg]$

Now dividing we get $\displaystyle \int{e^{-ax}\sin(bx)dx}=\frac{e^{-ax}\bigg[\frac{-\sin(bx)}{a}-\frac{b\cos(bx)}{a^2}\bigg]}{1+\frac{b^2}{a^2}}$

Multiplying top and bottom by $\displaystyle a^2$ we get

$\displaystyle \int{e^{-ax}\sin(bx)dx}=\frac{e^{-ax}\bigg[-\sin(bx)\cdot{a}-b\cos(bx)\bigg]}{a^2+b^2}$

and as was stated earlier $\displaystyle \int_0^{\infty}{e^{-ax}\sin(bx)dx}=\bigg[\frac{e^{-ax}\bigg[-\sin(bx)\cdot{a}-b\cos(bx)\bigg]}{a^2+b^2}\bigg]\bigg|_0^{\infty}=\frac{b}{a^2+b^2}$

Wow that took like 25 minutes to type ...I think thats right....o wait...now I know it is I found that integral in my books...I know the asnwer but I did the integral myself...hope I did it right

3. The whole thing is messy

I need nearly a minute to display correctly a page :/

More organization won't be extra luxury ^^'

4. Originally Posted by Moo
The whole thing is messy

I need nearly a minute to display correctly a page :/

More organization won't be extra luxury ^^'
Haha...I know it took me seven minutes to DO it...and like 30 minutes to type it lol

5. Another simpler yet not easy general case indefinite integral

$\displaystyle \int_0^{a}\frac{x^2+b^2}{x^2+a^2}dx$

First we use the +0 trick to obtian

$\displaystyle \int\frac{x^2+a^2+b^2-a^2}{x^2+a^2}dx=\int\frac{x^2+a^2}{x^2+a^2}+\int\f rac{b^2-a^2}{x^2+a^2}dx=\int{dx}+\int\frac{b^2-a^2}{x^2+a^2}dx$

Now by evaluating the first integral and factoring out a constant in the second we get

$\displaystyle x+(b^2-a^2)\int\frac{1}{x^2+a^2}dx$

where it becomes apparent the second integral is an arctan integral

$\displaystyle x+\frac{b^2-a^2}{a}arctan\bigg(\frac{x}{a}\bigg)$

Now by evaluating from 0 to a we get (for space's sake F(0)=0)

$\displaystyle \int_0^{a}\frac{x^2+b^2}{x^2+a^2}dx=a+\frac{b^2-a^2}{a}arctan\bigg(\frac{a}{a}\bigg)=a+\frac{b^2-a^2}{a}\cdot\frac{\pi}{4}$$\displaystyle =\frac{4a^2+(b^2-a^2)\pi}{4a}=\frac{(4-\pi)a^2+b^2\pi}{4a}=\frac{(4-\pi)a}{4}+\frac{b^2\pi}{4a}$

This is an integral where the simplification is two thirds of the work

6. Originally Posted by Mathstud28
This is another messy general case integral

We have $\displaystyle \int_0^{\infty}e^{-ax}sin(bx)dx$
Integration by parts it's a really waste of time. I do prefer (and I think teachers could show this to their students) to rewrite $\displaystyle e^{-ax}\sin(bx)=\text{Im }\Big\{e^{-x(a-bi)}\Big\}.$ From here we only have to compute $\displaystyle \text{Im }\frac1{a-bi}$ and we're done.

7. Originally Posted by Krizalid
Integration by parts it's a really waste of time. I do prefer (and I think teachers could show this to their students) to rewrite $\displaystyle e^{-ax}\sin(bx)=\text{Im }\Big\{e^{-x(a-bi)}\Big\}.$ From here we only have to compute $\displaystyle \text{Im }\frac1{a-bi}$ and we're done.
I am sorry...you have to remember that I am only in Calc II or the equivalent...coud you explain further...you are using someting like arg? I think Im() stands for image which I know to be synonomous with range? Please explain thoroughly if you have time....thanks so much in advance

8. Originally Posted by Mathstud28
I am sorry...you have to remember that I am only in Calc II or the equivalent...coud you explain further...you are using someting like arg? I think Im() stands for image which I know to be synonomous with range? Please explain thoroughly if you have time....thanks so much in advance
he is talking about interpreting the problem in the realm of complex analysis. "Im" stands for "imaginary part", that is, the coefficient of i in a complex number. don't worry about it if you have not done it... unless you are intrigued of course and want to learn more

9. Originally Posted by Jhevon
he is talking about complex analysis. "Im" stands for "imaginary part", that is, the coefficient of i in a complex number. don't worry about it if you have not done it... unless you are intrigued of course and want to learn more
Jhevon...its math...math I dont knwo...of course I am intrigued! Teach me teach me!

10. Originally Posted by Mathstud28
Jhevon...its math...math I dont knwo...of course I am intrigued! Teach me teach me!
It is perhaps better that you are taught by the master. i will allow Krizalid to offer a response. if he is not so inclined, then i will explain it.

for the record, I am sure you will enjoy complex analysis when you take it! it is a really fun part of math, ask anyone

11. Originally Posted by Jhevon
It is perhaps better that you are taught by the master. i will allow Krizalid to offer a response. if he is not so inclined, then i will explain it.

for the record, I am sure you will enjoy complex analysis when you take it! it is a really fun part of math, ask anyone
I know ...I am really excited...I want to do it now haha...I dont have the neccasary skill set yet do I?

12. There's no much here, for purely complex let $\displaystyle \sin x=\text{Im\,}(e^{ix})$ & $\displaystyle \cos x=\text{Re\,}(e^{ix}).$ (Where $\displaystyle \text{Re}$ denotes the real part.) Now apply this to that integral and use a well known fact (easy to prove) $\displaystyle \frac1x=\int_0^\infty e^{-xy}\,dy$ for $\displaystyle x>0.$

13. Originally Posted by Mathstud28
I know ...I am really excited...I want to do it now haha...I dont have the neccasary skill set yet do I?
actually, i think you do...

Originally Posted by Krizalid
There's no much here, for purely complex let $\displaystyle \sin x=\text{Im\,}(e^{ix})$ & $\displaystyle \cos x=\text{Re\,}(e^{ix}).$ (Where $\displaystyle \text{Re}$ denotes the real part.) Now apply this to that integral and use a well known fact (easy to prove) $\displaystyle \frac1x=\int_0^\infty e^{-xy}\,dy$ for $\displaystyle x>0.$

well you could give Euler's formula and explain how we treat a + ib as a constant and all that...

14. Originally Posted by Jhevon
actually, i think you do...

well you could give Euler's formula and explain how we treat a + ib as a constant and all that...
OHh...now I see sort of where this is coming from...I know Euler's formula to be $\displaystyle e^{ix}=\cos(x)+i\sin(x)$...so then obviously the real part (i.e. $\displaystyle Re[e^{ix}]$) is cos(x) and the imaginary part(i.e. $\displaystyle Im[e^{ix}]$) is sin(x)...ok so I understand that and the other part about the identity makes sense since $\displaystyle \int_0^{\infty}e^{-xy}dy=\frac{-e^{-xy}}{x}\bigg|_0^{\infty}=0-\frac{-1}{x}=\frac{1}{x}$...so now the cartesian form of this would be something to the affect that I use the complex argument or something on sin(x)? Could you explain from here?

15. Here just for fun I will derive all three of the "fun" trigonoetric integrals

Case $\displaystyle \sqrt{a^2-x^2}$

This one looks easy compared to the other two but lets start

$\displaystyle \int\sqrt{a^2-x^2}dx$

Let $\displaystyle x=a\sin(\theta)$
then $\displaystyle dx=a\cos(\theta)$

Then by substitution we have

$\displaystyle \int\sqrt{a^2-a\sin^2(\theta)}a\cos(\theta)d\theta$

$\displaystyle \int\sqrt{a^2(1-\sin^2(\theta))}a\cos(\theta)d\theta$

$\displaystyle \int\sqrt{a^2\cos^2(\theta)}a\cos(\theta)d\theta$

$\displaystyle a^2\int\cos^2(\theta)d\theta$

$\displaystyle \frac{a^2}{2}\int(1+\cos(2\theta))d\theta$...using$\displaystyle \cos^2(x)=\frac{1+\cos(2x)}{2}$

by integrating we get

$\displaystyle \frac{a^2}{2}\bigg[\theta+\frac{1}{2}\sin(2\theta)\bigg]$

using the identity $\displaystyle \sin(2x)=2\sin(x)\cos(x)$

we get $\displaystyle \frac{a^2}{2}\bigg[\theta+\sin(\theta)\cos(\theta)\bigg]$

Now using our intial substitution $\displaystyle x=a\sin(\theta)\Rightarrow{arcsin\bigg(\frac{x}{a} \bigg)=\theta}$

we get $\displaystyle \frac{a^2}{2}\bigg[arcsin\bigg(\frac{x}{a}\bigg)+sin\bigg(arsin\bigg( \frac{x}{a}\bigg)\bigg)\cos\bigg(arcsin\bigg(\frac {x}{a}\bigg)\bigg)\bigg]$

and using the fact that $\displaystyle cos\bigg(arcsin\bigg(\frac{x}{a}\bigg)\bigg)=\frac {\sqrt{a^2-x^2}}{a}$
and $\displaystyle \sin\bigg(arcsin\bigg(\frac{x}{a}\bigg)\bigg)=\fra c{x}{a}$

we see that the answer is $\displaystyle \frac{a^2}{2}\bigg[arcsin\bigg(\frac{x}{a}\bigg)+\frac{x\sqrt{a^2-x^2}}{a^2}\bigg]=\frac{arcsin\bigg(\frac{x}{a}\bigg)}{2a^2}+\frac{ 1}{2}x\sqrt{a^2-x^2}+C$

that took a while...so I will do the other two cases some other time...either tonight or tomorrow...I did this by hand without consulting a book...if anyone sees any errors please inform me by PM!!

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