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Help in Maths question (1 Viewer)

Thread starter Da_Bomb
Start date Dec 4, 2009

D

Da_Bomb

New Member

#1

Use the substitution x=3sintheatre, to verify that integral 1 divide by root of 9-xsquared = inverse sin ( x over 3 ) plus c.

Trebla

Administrator

Administrator

#2

$x=3\sin\theta\Rightarrow dx=3\cos\theta d\theta\\\\\int \dfrac{dx}{\sqrt{9-x^2}}=\int\dfrac{3\cos\theta}{\sqrt{9-9\sin^2\theta}}\,d\theta\\\\=\int\dfrac{3\cos\theta}{3\cos\theta}d\theta\\\\=\int d\theta\\\\= \theta + c\\\\$but $ x=3\sin\theta \Rightarrow \theta = \sin^{-1}\frac{x}{3}\\\\\therefore \int \dfrac{dx}{\sqrt{9-x^2}} = \sin^{-1}\frac{x}{3}+c$

kaz1

et tu

#3

Da_Bomb said:
Use the substitution x=3sintheatre, to verify that integral 1 divide by root of 9-xsquared = inverse sin ( x over 3 ) plus c.

lol, what a noob.

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