Integration (1 Viewer)

wantingtoknow · Jun 4, 2009

Help please, thanks.

∫dx/[x√(1-x²)]

MC Squidge · Jun 4, 2009

let x=sin@

wantingtoknow · Jun 4, 2009

MC Squidge said:
let x=sin@

How do I integrate cosecu? :/

gurmies · Jun 4, 2009

$\int \frac{dx}{x\sqrt{1-x^{2}}} \\\\ Let \,\, x=\sin\theta \\\\ dx=\cos\theta d\theta \\\\ \int \frac{\cos\theta d\theta}{\sin\theta.\cos\theta} = \int \csc\theta d\theta \\\\ = -\ln|\cot\theta+\csc\theta|+C \\\\ But \,\, \theta = \sin^{-1}x \\\\ Using \,\, right-angled \,\, triangle, \,\, \sin^{-1}x=\cot^{-1}(\frac{\sqrt{1-x^{2}}}{x})=\csc^{-1}(\frac{1}{x}) \\\\ \therefore I=-\ln|\frac{\sqrt{1-x^{2}}}{x}+\frac{1}{x}|+C=\ln|\frac{x}{\sqrt{1-x^{2}}+1}|+C$

NOTE: To integrate cosec@, multiply both top and bottom of cosec@/1 by (cot@+cosec@) and you end up with the derivative of the denominator in the numerator (natural log)

wantingtoknow · Jun 4, 2009

Thanks

I got it.

lolokay · Jun 4, 2009

a quicker way
$\int \frac{dx}{x\sqrt{1-x^2}}\\ =\int \frac{dx}{x^2\sqrt{(\frac{1}{x})^2-1}}$
use reverse chain rule/let u=1/x, as this is a standard integral

gurmies · Jun 4, 2009

lolokay said:
a quicker way
$\int \frac{dx}{x\sqrt{1-x^2}}\\ =\int \frac{dx}{x^2\sqrt{(\frac{1}{x^2})-1}}$
use reverse chain rule/let u=1/x, as this is a standard integral

I've had just about enough of you xD. No really, great solution.

MC Squidge · Jun 4, 2009

well dayumn!

Integration (1 Viewer)

wantingtoknow

Member

MC Squidge

BOS' Apex Predator

wantingtoknow

Member

gurmies

Drover

wantingtoknow

Member

lolokay

Active Member

gurmies

Drover

MC Squidge

BOS' Apex Predator

Users Who Are Viewing This Thread (Users: 0, Guests: 1)