a fitzy question (1 Viewer)

richz

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this question is from fitzy ex 26 (d) pg 156

prove that d/dx (xsin^-1 (x)) = sin^-1(x) + x/(root(1-x^2). hence find the primitive of sin^-1(x) and show that: the integral between .5 and 0 (sin^-1 (x)) = (pi/12) + (root3/2) -1

thnx
 

FinalFantasy

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xrtzx said:
this question is from fitzy ex 26 (d) pg 156

prove that d/dx (xsin^-1 (x)) = sin^-1(x) + x/(root(1-x^2). hence find the primitive of sin^-1(x) and show that: the integral between .5 and 0 (sin^-1 (x)) = (pi/12) + (root3/2) -1

thnx
"prove that d/dx (xsin^-1 (x)) = sin^-1(x) + x/(root(1-x^2). "
d/dx (xsin^-1 (x))=sin^-1 (x)+x\sqrt(1-x²) product rule..

"hence find the primitive of sin^-1(x)"

d/dx (xsin^-1 (x))=sin^-1 (x)+x\sqrt(1-x²)
sin^-1 (x)=d\dx(xsin^-1(x))-x\sqrt(1-x²)
integrate both sides
primitive of sin^-1 (x)=xsin^-1 (x)-int. x\sqrt(1-x²) dx


consider I=int. x\sqrt(1-x²) dx
let u=1-x²
du\dx=-2x
.: I=-1\2 int. du\u^(1\2)
=-u^(1\2)=-sqrt(1-x²)

primitive of sin^-1 (x)=xsin^-1 (x)-int. x\sqrt(1-x²) dx
primitive of sin^-1 (x)=xsin^-1 (x)+sqrt(1-x²)+C
 

FinalFantasy

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i like it how u always bold my username
hahaha
do u like Final Fantasy- the game?
 

who_loves_maths

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Originally Posted by FinalFantasy
i like it how u always bold my username
hahaha
do u like Final Fantasy- the game?
actually it's ironic, cause i've never played it.
no i bold or put in italics everyone's nick that i mention in any of my posts as a sign of respect... just something i'm used to. and also so as not to confuse nick's with normal words and phrases, etc... seeing how some nick's are not actually distinguishable from the vernacular.

but in this case, you had priority in posting the solution to this problem, so bolding your name is highlighting my acknowledgement of your antecedent solution to my "tidying-things-up" post ;)
 

FinalFantasy

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who_loves_maths said:
actually it's ironic, cause i've never played it.
no i bold or put in italics everyone's nick that i mention in any of my posts as a sign of respect... just something i'm used to. and also so as not to confuse nick's with normal words and phrases, etc... seeing how some nick's are not actually distinguishable from the vernacular.

but in this case, you had priority in posting the solution to this problem, so bolding your name is highlighting my acknowledgement of your antecedent solution to my "tidying-things-up" post ;)
hahaha, but dun u find it annoying typing all those codes in all the time
like having to type [bold /] and stuff
 

Slidey

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If you don't want to use integration by parts, you can use the rectangle method.
 

Slidey

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LOL. I'm pretty sure I've mentioned it before.

Int sin^-1(x) dx from 0 to 1/2

y=arcsinx
x=siny
0=siny -> y=0
1/2=siny -> y=pi/6

Int siny dy from 0 to pi/6
=sqrt(3)/2 - 1

However, since that was dy, it went around the y access, and is not directly the area, we want. So we form the rectangle which would encompass exactly both the area around the x and the area around the y axes. This rectangle has dimensions (b-a)*(f(b)-f(a)) which in this case is (1/2 - 0)*(pi/6 - 0)=pi/12
Now take the area around the y-axis from this rectangle's area to get the area around the x-axis: pi/12-sqrt(3)/2+1, or Int sin^-1(x) dx from 0 to 1/2
 

Slidey

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Don't worry, there should be a diagram to help. But I'm lazy.

Refer to page 153 of fitz 3u for diagrams.
 

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