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Evaluate
∫ (1÷(1+cos×(x))
Why we cannot rationalise it with 1-cosx
MZk
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I perform rationalization and ans is wrong
MZk
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What did you get?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi Zeeshan 01
What do you mean by rationalise here?
eg.
This is called rationalisation because the irrational denominator has been made into a rational.
But cosine(x) isn't an irrational. ??
To do the integral you can make use of
This makes a function that is directly integrable.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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You can use the tangent half-angle substitution.
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I know half angle but why we not rationaliz
How???
But cosine(x) isn't an irrational. ??
Why we cannot multiply and divide by 1-cos (x)
MZk
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Why we cannot multiply and divide by 1-cos (x)
No one is saying you can't do that, you can do that if you want. Post your calculation here and tell us what you think about the integral.
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Ok
MZk
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By doing rationalization
1-cosx÷sin (x)^2
Then
1÷sin^2 (x) -cos (x)÷
MZk
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You are missing something in the last line.
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1÷sin^2 (x) -cos (x)÷sin^2 (x)
MZk
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I solve ans is cosecx-cot x+c
MZk
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At first I thought this was not correct as my answer was different. But both results have the same graph so then I checked the identity and we have the same.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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I would recommend the tangent half-angle substitution for integrals of this type.
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I also know half angle but why not this
t both results have the same graph ????? Where you plotted graph
Then I checked the identity and we have the same. smile
Which identity???
MZk
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By doing rationalization
1-cosx÷sin (x)^2 . . . . . . This isn't correct, because you didn't use grouping symbols.
[1 - cos(x)]÷[sin(x)]^2
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I know
MZk
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hi Zeeshan 01
I did this integral a different way.
So my first thought when I saw your answer was ???
So I plotted the two functions together on a graph. The second plot fitted exactly over the first. To show this more clearly I have offset my graph by 0.3 along the axis so you can see both graphs.
Then I checked to see if I could prove they are the same by using identity methods:
So we have the same answer.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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Ok no one can wrong my method !!!
MZk
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hi Zeeshan 01
Your method and answer are both good. With any integration, if your answer differentiates back to the original question, then it is a good answer.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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I know !!!! But thx
And this question check this
∫ 1÷(x^2+4x+13)dx
How to solve this??
MZk
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Try completing the square and then using a trig substitution.
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What is trig substitution
When I do 1÷((x+2)^2+3^2)
It's tan formula
MZk
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Plese show me
MZk
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