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Size:  518.5 KB for this question I don't get his the limits are worked out. Attachment 618360618362 I know that I wasn't supposed to use the limits 1/2 and 0 but I don't get how pi/6 is worked out.

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    (Original post by coconut64)
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    LImits for x are 0 and \frac{1}{2}. So with the substitution x = \sin \theta, you substitute in the limits of the integral and solve for \theta to obtain the new limits.

    \frac{1}{2} = \sin \theta

    \theta = \sin ^{-1} (\frac{1}{2})

    \theta = \frac{\pi}{6}

    Similarly, when x=0, theta = 0.
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    (Original post by coconut64)
    Name:  1486678658454-2143366265.jpg
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Size:  518.5 KB for this question I don't get his the limits are worked out. Attachment 618360618362 I know that I wasn't supposed to use the limits 1/2 and 0 but I don't get how pi/6 is worked out.

    Thanks
    Once you make the substition you should either change the limits, as explained above, or write x = 0 and x = 1/2 on the integral.

    Then after integrating either substitute back or change limits then, You must not 'mix up'and integration of u and x.

    What you have written is incorrect ...
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    (Original post by Muttley79)
    Once you make the substition you should either change the limits, as explained above, or write x = 0 and x = 1/2 on the integral.

    Then after integrating either substitute back or change limits then, You must not 'mix up'and integration of u and x.

    What you have written is incorrect ...
    I don't get what you mean by the bit underlined. Initially I subbed sin theta straight into the equation , rather than integrating the equation involving x. I did replace dx with d theta. I know where I have gone wrong with the limits..

    Thanks
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    (Original post by coconut64)
    I don't get what you mean by the bit underlined. Initially I subbed sin theta straight into the equation , rather than integrating the equation involving x. I did replace dx with d theta. I know where I have gone wrong with the limits..

    Thanks
    Meaning you can ignore the limits until after you integrate. At which point, you may choose to convert your expression into terms of x and apply the same limits, or instead you can change the limits to work with the expression in terms of \theta

    So at the stage of \displaystyle [\tan(\theta)] you could've either converted into \displaystyle [f(x)]_0^{\frac{1}{2}} (which doesn't really apply to this question well) OR \displaystyle [f(\theta)]_{\theta(0)}^{\theta(\frac{1}{2}  )} but not a mix as you have there.

    By \theta(x) I denote \arcsin(x)
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    (Original post by RDKGames)
    Meaning you can ignore the limits until after you integrate. At which point, you may choose to convert your expression into terms of x and apply the same limits, or instead you can change the limits to work with the expression in terms of \theta

    So at the stage of \displaystyle [\tan(\theta)] you could've either converted into \displaystyle [f(x)]_0^{\frac{1}{2}} (which doesn't really apply to this question) OR \displaystyle [f(\theta)]_{\theta(0)}^{\theta(\frac{1}{2}  )} but not a mix as you have there.

    By \theta(x) I denote \arcsin(x)
    So I just basically need to change the limit? Since I have already have my f(x) in place, which is tan theta. So are you suggesting that this should have been [tan theta] pi/6 on top and 0 at the bottom? Thanks
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    (Original post by coconut64)
    So I just basically need to change the limit? Since I have already have my f(x) in place, which is tan theta. So are you suggesting that this should have been [tan theta] pi/6 on top and 0 at the bottom? Thanks
    That is not f(x), that is f(\theta).

    And yes that is what everyone on this thread is suggesting - the change in limits.
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    (Original post by RDKGames)
    That is not f(x), that is f(\theta).

    And yes that is what everyone on this thread is suggesting - the change in limits.
    What is f(x)? X has been replaced by sin theta already, I don't understand what you mean by suddenly change it from f theta back to f(x), The equation has been simplified to sec^2 x.... If I do tan(pi/6) - tan(0), this gives me the answer which is 1/3 (root 3)
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    (Original post by coconut64)
    What is f(x)? X has been replaced by sin theta already, I don't understand what you mean by suddenly change it from f theta back to f(x), The equation has been simplified to sec^2 x....
    With your particular question, f(x) is just \tan(\theta) in terms of x (or \sin(\theta)) which would be \displaystyle \tan(\theta)= \frac{\sin(\theta)}{\sqrt{1-\sin^2(\theta)}}= \frac{x}{\sqrt{1-x^2}} for some \theta and replacing this in your integrated expression is too messy.

    I'll give you a simple example concerning the two approaches.

    \displaystyle 3 \int_0^1 2x(x^2+1)^2 .dx then u=x^2+1 \Rightarrow dx=\frac{du}{2x}

    So we have 3 \int u^2 .du = u^3

    Now we can either convert it back to x and use the original limits in which case we have \displaystyle [(x^2+1)^3]_0^1 (this approach is not recommended for your Q) OR we can convert the limits so we have \displaystyle [u^3]_1^2. Both of which give us the same answer.
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    (Original post by RDKGames)
    With your particular question, f(x) is just \tan(\theta) in terms of x (or \sin(\theta)) which would be \displaystyle \tan(\theta)= \frac{\sin(\theta)}{\sqrt{1-\sin^2(\theta)}}= \frac{x}{\sqrt{1-x^2}} for some \theta and replacing this in your integrated expression is too messy.

    I'll give you a simple example concerning the two approaches.

    \displaystyle 3 \int_0^1 2x(x^2+1)^2 .dx then u=x^2+1 \Rightarrow dx=\frac{du}{2x}

    So we have 3 \int u^2 .du = u^3

    Now we can either convert it back to x and use the original limits in which case we have \displaystyle [(x^2+1)^3]_0^1 (this approach is not recommended for your Q) OR we can convert the limits so we have \displaystyle [u^3]_1^2. Both of which give us the same answer.
    To simplify this, are you suggesting my method takes too long?
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    (Original post by coconut64)
    To simplify this, are you suggesting my method takes too long?
    No, I'm only explaining what Muttley has stated when he said "You must not 'mix up'and integration of u and x" because that's what you've done with your method.
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    (Original post by RDKGames)
    No, I'm only explaining what Muttley has stated when he said "You must not 'mix up'and integration of u and x" because that's what you've done with your method.
    Sorry to drag on to this forever but I am more confused than ever.

    Mutterfly said I shouldn't have mixed up u and x when integrating. I get where he is coming from, because I did not change the limits 0.5 &0, although I am integrating with respect with d theta. I understand this bit.

    But you are suggesting that tan theta is equivalent to the orginal equation we are given x/ (1-x^2)^1.5 , but tan theta is obtained after I have integrated sec^2 x. Surely sec^2x is equal to sin theta / (1- x^2)^1.5. ThanksAttachment 618834 thanks
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    (Original post by coconut64)
    Sorry to drag to this forever but I am more confused than ever.

    Mutterfly said I shouldn't have mixed up u and x when integrating. I get where he is coming from, because I did not change the limits 0.5 &0, although I am integrating with respect with d theta. I understand this bit.

    But you are suggesting that tan theta is equivalent to the orginal equation we are given x/ (1-x^2)^1.5 , but tan theta is obtained after I have integrated sec^2 x. Surely sec^2x is equal to sin theta / (1- x^2)^1.5. Thanks
    I am not claiming it is equal to the original. The original is \frac{1}{\sqrt{1-x^2}} whereas I claim that \tan(\theta) = \frac{x}{\sqrt{1-x^2}} for some \theta - but this isn't a true enough statement as it's not true for all \theta hence why this approach doesn't work very well. I have used a simpler integral to simply display the two approaches - which you now seem to understand.
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    (Original post by RDKGames)
    I am not claiming it is equal to the original. The original is \frac{1}{\sqrt{1-x^2}} whereas I claim that \tan(\theta) = \frac{x}{\sqrt{1-x^2}} for some \theta - but this isn't a true enough statement as it's not true for all \theta hence why this approach doesn't work very well. I have used a simpler integral to simply display the two approaches - which you now seem to understand.

    Since x=\sin(\theta) you can
    I have edited my post. Are you sure? The original equation is to the power of 1.5, not 1/2. Thanks
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    (Original post by coconut64)
    I have edited my post. Are you sure? The original equation is to the power of 1.5, not 1/2. Thanks
    Either way, yes I am sure because in either case I did not state that \tan(\theta) is equal to it as you claim.
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    (Original post by RDKGames)
    Either way, yes I am sure because in either case I did not state that [tex]\tan(\theta)[tex] is equal to it as you claim.
    I have checked, the original equation is to the power of 1.5, not 0.5. But I appreciate your help though.
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    (Original post by coconut64)
    I have checked, the original equation is to the power of 1.5, not 0.5. But I appreciate your help though.
    Almost (*) nothing RDK has posted has anything to do with the form of the original equation, so this has nothing to do with his points and explanations.

    (*) AFAIK, literally the only thing he's said about the original equation is "The original is \frac{1}{\sqrt{1-x^2}}". Yes this isn't correct, but since he only quoted it to say "I'm not talking about the original integral I'm talking about  \frac{x}{\sqrt{1-x^2}}" it doesn't invalidate anything else he said. He's still not talking about the original equation, and it still is a different equation from  \frac{x}{\sqrt{1-x^2}}
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    (Original post by coconut64)
    I have checked, the original equation is to the power of 1.5, not 0.5. But I appreciate your help though.
    Just to clarify further, you have defined x=\sin(\theta)

    If you wish to keep your limits the same, then you must express \tan(\theta) in terms of \sin(\theta) thus express it in terms of x. But you cannot define \tan(\theta) in terms of \sin(\theta) for all \theta. You can sure try, and it works, but it doesn't hold for all theta hence why it breaks down when you come to use it and why the approach of simply changing the limits is the better option. To do this you would have to go \displaystyle \tan(\theta)=\frac{\sin(\theta)}  {\cos(\theta)} == \frac{\sin( \theta) }{\sqrt{\cos^2(\theta)}}=\frac{ \sin(\theta)}{\sqrt{1-\sin^2(\theta)}}=\frac{x}{\sqrt{  1-x^2}}

    I have indicated the step where it breaks down by '=='

    As said by DFranklin above, it has nothing to do with the original expression.
 
 
 
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