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..
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 θ
So at the stage of [tan(θ)] you could've either converted into [f(x)]021 (which doesn't really apply to this question well) OR [f(θ)]θ(0)θ(21) but not a mix as you have there.
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 θ
So at the stage of [tan(θ)] you could've either converted into [f(x)]021 (which doesn't really apply to this question) OR [f(θ)]θ(0)θ(21) but not a mix as you have there.
By θ(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
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(θ).
And yes that is what everyone on this thread is suggesting - the change in limits.
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 X has been replaced by sin theta already, I don't understand what you mean by suddenly change it from f theta back to f(x), The equation has been simplified to sec^2 x....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)
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(θ) in terms of x (or sin(θ)) which would be tan(θ)=1−sin2(θ)sin(θ)=1−x2x for some θ and replacing this in your integrated expression is too messy.
I'll give you a simple example concerning the two approaches.
3∫012x(x2+1)2.dx then u=x2+1⇒dx=2xdu
So we have 3∫u2.du=u3
Now we can either convert it back to x and use the original limits in which case we have [(x2+1)3]01 (this approach is not recommended for your Q) OR we can convert the limits so we have [u3]12. Both of which give us the same answer.
With your particular question, f(x) is just tan(θ) in terms of x (or sin(θ)) which would be tan(θ)=1−sin2(θ)sin(θ)=1−x2x for some θ and replacing this in your integrated expression is too messy.
I'll give you a simple example concerning the two approaches.
3∫012x(x2+1)2.dx then u=x2+1⇒dx=2xdu
So we have 3∫u2.du=u3
Now we can either convert it back to x and use the original limits in which case we have [(x2+1)3]01 (this approach is not recommended for your Q) OR we can convert the limits so we have [u3]12. Both of which give us the same answer.
To simplify this, are you suggesting my method takes too long?
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.
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. Thanks
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 1−x21 whereas I claim that tan(θ)=1−x2x for some θ - but this isn't a true enough statement as it's not true for all θ 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.
I am not claiming it is equal to the original. The original is 1−x21 whereas I claim that tan(θ)=1−x2x for some θ - but this isn't a true enough statement as it's not true for all θ 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(θ) you can
I have edited my post. Are you sure? The original equation is to the power of 1.5, not 1/2. Thanks
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 1−x21". 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 1−x2x" 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 1−x2x
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(θ)
If you wish to keep your limits the same, then you must express tan(θ) in terms of sin(θ) thus express it in terms of x. But you cannot define tan(θ) in terms of sin(θ) for all θ. 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 tan(θ)=cos(θ)sin(θ)==cos2(θ)sin(θ)=1−sin2(θ)sin(θ)=1−x2x
I have indicated the step where it breaks down by '=='
As said by DFranklin above, it has nothing to do with the original expression.