Author Topic: Setting out integration by substitution (Read 1066 times) Tweet Share

NE2000 · « **on:** October 31, 2009, 02:13:39 pm »

It is never too late to ask right? This stuff only occurred to me during the exam yesterday. Otherwise I've always been comfortable with it. Thought I'll get it cleared up.

In substitution I heard you have to show working out every time to do it saying (for example):

$\frac {du}{dx} dx = du$

I thought that was just assumed?

Secondly, is this setting out appropriate?

$\int \frac {6x}{2x^2+1} dx$

$let u = 2x^2 + 1$

$\frac {du}{dx} = 4x$

$\frac {3}{2} \frac {du}{dx} = 6x$

$\frac {3}{2} \int \frac {1}{u} du$

I've been told that the 'proper' setting out is:

$\int \frac {6x}{u} \frac {du}{4x}$

and working from there.

Just wanted to ensure I was doing nothing mathematically wrong.

EDIT: I may sound paranoid, but I just don't want to lose unexpected marks out of

GerrySly · « **Reply #1 on:** October 31, 2009, 02:18:36 pm »

This is how I've been told to set it out

$\begin{align*}Let u&=2x^2+1\\<br />\therefore \frac{du}{dx}&=4x\\<br />\implies dx&=\frac{du}{4x}\end{align*}$

I do that on the side, then just make the substitution and make my integral...

$\begin{align*}\int \frac{6x}{2x^2+1}.dx&=\int \frac{6x}{u}\frac{du}{4x}\\&=\frac{3}{2}\int\frac{1}{u}.du\end{align*}$

/0 · « **Reply #2 on:** October 31, 2009, 02:45:27 pm »

I've done it like this:

Let $u = 2x^2+1$

$du = 4xdx$

$\int \frac{6x}{2x^2+1}dx = \frac{3}{2}\int \frac{4xdx}{2x^2+1} = \frac{3}{2}\int \frac{du}{u}$

TrueTears · « **Reply #3 on:** October 31, 2009, 03:51:51 pm »

Quote from: /0 on October 31, 2009, 02:45:27 pm

I've done it like this:

Let $u = 2x^2+1$

$du = 4xdx$

$\int \frac{6x}{2x^2+1}dx = \frac{3}{2}\int \frac{4xdx}{2x^2+1} = \frac{3}{2}\int \frac{du}{u}$

Yeap I always treat it as differentials like /0

Mao · « **Reply #4 on:** October 31, 2009, 11:00:42 pm »

do a substitution like a real substitution:

$I = \int \cos(x) \sin^2(x)\; dx$

Let $u = \sin (x) \implies \frac{du}{dx} = \cos (x) \implies dx = \frac{du}{\cos(x)}$

$\implies I = \int \cos(x) \cdot u^2 \cdot \frac{du}{\cos (x)}$

d0minicz · « **Reply #5 on:** October 31, 2009, 11:12:31 pm »

do you HAVE to show that amount of working for them not to take marks off

dcc · « **Reply #6 on:** October 31, 2009, 11:16:42 pm »

It can't hurt to do what Mao did (which would be similar to what I would write). At worst, you've spent 10 seconds ensuring that you don't make a stupid mistake.

dejan91 · « **Reply #7 on:** October 31, 2009, 11:54:10 pm »

I do it this way:

$I = \int \cos(x)( \sin^2(x)\; dx$
Let $u=\sin(x) \implies \frac{du}{dx} = \cos(x)$
$\therefore I = \int u^2\frac{du}{dx}\; dx \implies I = \int u^2\; du$

Search the forums now!

We have moved!

Author Topic: Setting out integration by substitution (Read 1066 times) Tweet Share

NE2000

Setting out integration by substitution

GerrySly

Re: Setting out integration by substitution

/0

Re: Setting out integration by substitution

TrueTears

Re: Setting out integration by substitution

Mao

Re: Setting out integration by substitution

d0minicz

Re: Setting out integration by substitution

dcc

Re: Setting out integration by substitution

dejan91

Re: Setting out integration by substitution

Recent Posts

User:
Password: