Integration (1 Viewer)

cutemouse · Apr 30, 2009

Hi, could someone help me? Thanks

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

Drongoski · Apr 30, 2009

1 + cos^2 x = (3+cos2x)/2

use half-angle formula.

lolokay · Apr 30, 2009

$\int \frac{1}{sin^2x + 2cos^2x}dx\\ \int \frac{sec^2x}{tan^2x + 2}dx$
which is then a standard integral

Drongoski · Apr 30, 2009

lolokay said:
$\int \frac{1}{sin^2x + 2cos^2x}dx\\ \int \frac{sec^2x}{tan^2x + 2}dx$
which is then a standard integral

Hey . . . that's neat!

Timothy.Siu · Apr 30, 2009

lolokay said:
$\int \frac{1}{sin^2x + 2cos^2x}dx\\ \int \frac{sec^2x}{tan^2x + 2}dx$
which is then a standard integral

nice work.

cutemouse · May 1, 2009

Hi, I have another question. Could someone please help me? Thanks

$\int \frac{dx}{cos^2(x/2)-sin^2(x/2)}$

Thanks

Trebla · May 1, 2009

jm01 said:
Hi, I have another question. Could someone please help me? Thanks

$\int \frac{dx}{cos^2(x/2)-sin^2(x/2)}$

Thanks

The denominator becomes cos x (using half angle formula), so its simply the integral of sec x which is ln (sec x + tan x) + c (obtained by multiplying sec x by (sec x + tan x)/(sec x + tan x) and recognising numerator is derivative of the denominator...)

tommykins · May 1, 2009

lolokay said:
$\int \frac{1}{sin^2x + 2cos^2x}dx\\ \int \frac{sec^2x}{tan^2x + 2}dx$
which is then a standard integral

hhahha you can't be fucking serious mate.

lolokay · May 1, 2009

tommykins said:
hhahha you can't be fucking serious mate.

..huh?

gurmies · May 1, 2009

Trebla said:
The numerator becomes cos x (using half angle formula), so its simply the integral of sec x which is ln (sec x + tan x) + c (obtained by multiplying sec x by (sec x + tan x)/(sec x + tan x) and recognising numerator is derivative of the denominator...)

Should be denominator

How knit-picky am I lol..

tommykins · May 1, 2009

lolokay said:
..huh?

genius is all

cutemouse · May 1, 2009

Hi, I got another question...

$\int \frac{dx}{sinx cosx}$

Thanks

lolokay · May 1, 2009

$\int \frac{sin^2x + cos^2x}{sinxcosx}dx\\ \int \frac{sinx}{cosx} + \frac{cosx}{sinx}dx\\ -ln|cosx| + ln|sinx|\\ = ln|tanx| + C$

or

$\frac{1}{sinxcosx} = \frac{sec^2x}{tanx}$
which gets you it in one step, but hard to spot

cutemouse · May 1, 2009

sorry about I've got another one...

$\int \frac{(x+1)dx}{(x^2-3x+2)^\frac{1}{2}}$

Thanks

Timothy.Siu · May 1, 2009

jm01 said:
sorry about I've got another one...

$\int \frac{(x+1)dx}{(x^2-3x+2)^\frac{1}{2}}$

Thanks

change the top to make it the derivative of the bottom, and then with the other part, complete the square and its a standard integral

gurmies · May 1, 2009

cutemouse · May 2, 2009

Thanks gurmies...

I have a little problem again here...

The question was:

$\int (9-x^2)^\frac{-1}{2} dx$

The answer in the back was:

$\frac{1}{2}( x \sqrt{9-x^2} + 9sin^{-1}\frac{x}{3})$

I put it through Wolfram Mathematica and got:

$\frac{1}{2} x \sqrt{9+x^2}+\frac{9}{2} \text{ArcSinh}\left[\frac{x}{3}\right]$

I thought the answer would just be:
$sin^{-1}\frac{x}{3}$

Which is the correct answer, and what is ArcSinh??

Thanks

gurmies · May 2, 2009

$\int \frac{dx}{9-x^{2}}$

$Let \,\, x = 3\sin\theta$

$dx = 3\cos\theta d\theta$

$\int \frac{3\cos\theta d\theta}{9-9\sin^{2}\theta} = \frac{1}{3}\int \sec\theta d\theta$

$= \frac{1}{3}\ln|\sec\theta + \tan\theta| + C$

$But \,\, \theta = \sin^{-1}\frac{x}{3}$

$So \,\, \frac{1}{3}\ln|\sec(\sin^{-1}\frac{x}{3})+\tan(\sin^{-1}\frac{x}{3})|$

$\sin^{-1}\frac{x}{3}=\sec^{-1}\frac{3}{\sqrt{9-x^{2}}} \,\,\,\,\,\,\, Using \,\, right-angled \,\, triangle.$

$Similarly \,\, \sin^{-1}\frac{x}{3}=\tan^{-1}\frac{x}{\sqrt{9-x^{2}}}$

$\therefore \frac{1}{3}\ln|\frac{3}{\sqrt{9-x^{2}}}+\frac{x}{\sqrt{9-x^{2}}}| + C$

I have no practise with doing 4 unit substitutions because we haven't done that at school yet. I took a guess and it looks very wrong, so please judge me not on this attempt xD. To answer your questions:

1: sinh and cosh are hyperbolic functions - our teacher once told us that, but i'm not too sure what it means

2: Why is it not simply arcsin(x/3)? Because there is no root over 9-x^2

Round 2 for fun:

$\frac{1}{(9-x)^{2}} = \frac{1}{(9-x)(9+x)}$

$= \frac{1}{18(9-x)}+\frac{1}{18(9+x)}$

$\therefore \int \frac{dx}{9-x^{2}} = \frac{1}{18}\int (\frac{1}{9-x}+\frac{1}{9+x})dx$

$= \frac{1}{18}(-\ln|9-x|+\ln|9+x|)+C$

$= \frac{1}{18}\ln|\frac{9+x}{9-x}| + C$

Nothing like your answer thus far =(

cutemouse · May 2, 2009

sorry gurmies.. wrong question.. there was supposed to be a square root.

gurmies · May 2, 2009

ZOMOGOD -.- LOL

I believe you typed something wrong into the integrator. Try type: 1/(9-x^2)^(1/2)

http://integrals.wolfram.com/index.jsp?expr=1/(9-x^2)^(1/2)&random=false

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