Integration by substitution (2 Viewers)

sinophile · Mar 10, 2009

Find $\int t/sqrt(1-t^2)dt, using, u=1-t^2$

H4rdc0r3 · Mar 10, 2009

fail. why would you need to use u, + you forgot dt.

$-\frac{1}{2}ln(1-t^{2})$ + C

sinophile · Mar 10, 2009

thats not what the textbook solution says.

AlexJB · Mar 10, 2009

$\int \frac{t}{1-t^2} dt$

$Let u = 1-t^2$

$Then, du = -2t$

$\therefore \int \frac{t}{1-t^2} dt = -\frac{1}{2}\int \frac{du}{u} dt$

$= -\frac{1}{2} lnu + c$

$= -\frac{1}{2} ln(1-t^2) + c$

wantingtoknow · Mar 10, 2009

sinophile said:
thats not what the textbook solution says.

Is it because you did a typo? Is it dt instead of t as the numerator?

Timothy.Siu · Mar 10, 2009

wantingtoknow said:
Is it because you did a typo? Is it dt instead of t as the numerator?

then thats 4unit integration

sinophile · Mar 10, 2009

I didnt include the dt, sorry.

thenuker6 · Mar 11, 2009

Timothy.Siu said:
then thats 4unit integration

It's actually 3 unit as they give u the substitution

Drongoski · Mar 11, 2009

I find the practice of NSW HSC in providing the suggested substitution for integration makes this part of 3U really wimpish.

jet · Mar 11, 2009

They didn't give the substitution for one question last year in the 3u paper. That really really surprised me.

Timothy.Siu · Mar 11, 2009

thenuker6 said:
It's actually 3 unit as they give u the substitution

ur an idiot, learn to read

addikaye03 · Mar 11, 2009

jetblack2007 said:
They didn't give the substitution for one question last year in the 3u paper. That really really surprised me.

yer same, it was Q1 aswell i believe. Usually very very straight forward, no tricks.

Alot of poorer MX1 students would have seen the Sin^2x and straight away gone:
Cos2x=1-2sin^2x
...
sin^2x=1/2(1-cos2x) etc etc

When do the BOS notes for 2008 come out?

sinophile · Mar 11, 2009

isnt anyone going to answer my question ;_;

kwabon · Mar 11, 2009

AlexJB said:
$\int \frac{t}{1-t^2} dt$

$Let u = 1-t^2$

$Then, du = -2t$

$\therefore \int \frac{t}{1-t^2} dt = -\frac{1}{2}\int \frac{du}{u} dt$

$= -\frac{1}{2} lnu + c$

$= -\frac{1}{2} ln(1-t^2) + c$

thats how u do it mate

Njn · Mar 11, 2009

AlexJB said:
$\int \frac{t}{1-t^2} dt$

$Let u = 1-t^2$

$Then, du = -2t$

$\therefore \int \frac{t}{1-t^2} dt = -\frac{1}{2}\int \frac{du}{u} dt$

$= -\frac{1}{2} lnu + c$

$= -\frac{1}{2} ln(1-t^2) + c$

Yeah, that's how. Maybe you don't know the formula you need to use ? if you don't know logarithms and exponentials just follow it =)

http://img24.imageshack.us/my.php?image=submathsworking.jpg

Edit: I just noticed equation editor xD

Timothy.Siu · Mar 11, 2009

S f'(x)/f(x) dx = ln(f(x)

sinophile · Mar 11, 2009

But the textbook's solution says -sqrt(1-t^2)+c. This is from the fitzpatrick 3-u book btw.

Timothy.Siu · Mar 11, 2009

sinophile said:
But the textbook's solution says -sqrt(1-t^2)+c. This is from the fitzpatrick 3-u book btw.

then the original question was supposed to have a square root

sinophile · Mar 11, 2009

Timothy.Siu said:
then the original question was supposed to have a square root

.. FUCK

edited my original post

Njn · Mar 11, 2009

Timothy.Siu said:
then the original question was supposed to have a square root

sinophile said:
FUCK

Lol.

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