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    Hi everyone,

    Suppose I've to integrate  \frac{2tanx}{tan2x} using double angle identities, I'm guessing the id I need is \tan2A= \frac {2tanA}{1-tan^2A}?

    My problem is I'm not sure how to get what I need to integrate in the form of the identity. If I let A=x for the numerator, the denominator is way off.
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    Let's look at this

    \tan(2x)=\dfrac{2\tan(x)}{1-\tan^{2}(x)}

    \dfrac{2\tan(x)}{\tan(2x)}= \dfrac{2\tan(x)}{\dfrac{2\tan(x)  }{1-\tan^{2}(x)}}

    Dividing top and bottom by 2\tan(x) and multiplying by 1-\tan^{2}(x) will give you 1-\tan^{2}(x) to integrate, this can be done easily with use of \sec^{2}(x)=1+\tan^{2}(x)
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    (Original post by Robbie242)
    Let's look at this

    \tan(2x)=\dfrac{2\tan(x)}{1-\tan^{2}(x)}

    \dfrac{2\tan(x)}{\tan(2x)}= \dfrac{2\tan(x)}{\dfrac{2\tan(x)  }{1-\tan^{2}(x)}}

    Dividing top and bottom by 2\tan(x) and multiplying by 1-\tan^{2}(x) will give you 1-\tan^{2}(x) to integrate, this can be done easily with use of \sec^{2}(x)=1+\tan^{2}(x)

    Thanks very much for your help. It's a little complicated for me!

    If I follow your steps so I can understand it, I don't arrive at the same answer (I know you said I can just multiply  sec^2x but I wanted to it step by step to help me understand).

    Dividing top and bottom by 2tanx:

    \frac{2tanx}{2tanx} \div \frac{tan2x}{2tanx} = \frac{2tanx}{2tanx} \times \frac {2tanx}{tan2x} = \frac{4tan^2x}{(2tanx)(tan2x)}

    Multiplying top and bottom by 1-tan^2x


    \frac{4tan^2x}{(2tanx)(tan2x)} \times \frac{1-tan^2x}{1-tan^2x}  =   \frac{4tan^2x - 4tan^4x}{(2tanx)(tan2x)(-2tan^3x)(-tan^32x)}  A real mess! What did I do wrong?
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    (Original post by marcsaccount)
    Thanks very much for your help. It's a little complicated for me!

    If I follow your steps so I can understand it, I don't arrive at the same answer (I know you said I can just multiply  sec^2x but I wanted to it step by step to help me understand).

    Dividing top and bottom by 2tanx:

    \frac{2tanx}{2tanx} \div \frac{tan2x}{2tanx} = \frac{2tanx}{2tanx} \times \frac {2tanx}{tan2x} = \frac{4tan^2x}{(2tanx)(tan2x)}

    Multiplying top and bottom by 1-tan^2x


    \frac{4tan^2x}{(2tanx)(tan2x)} \times \frac{1-tan^2x}{1-tan^2x}  =   \frac{4tan^2x - 4tan^4x}{(2tanx)(tan2x)(-2tan^3x)(-tan^32x)}  A real mess! What did I do wrong?
    You seem to have flipped things a bit

    What I'd do is express your integral using the addition formulae before we divide \dfrac{2\tan(x)}{\tan(2x)}= \dfrac{2\tan(x)}{\dfrac{2\tan(x)  }{1-\tan^{2}(x)}}

    Now dividing top and bottom by 2\tan(x) will give the expression

    \dfrac{\frac{2\tan(x)}{2\tan(x)}  }{\frac{2\tan(x)}{2\tan(x)(1-\tan^{2}(x))}}=...
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    ...but I don't understand the step you've taken here

    (Original post by Robbie242)
    \dfrac{2\tan(x)}{\tan(2x)}= \dfrac{2\tan(x)}{\dfrac{2\tan(x)  }{1-\tan^{2}(x)}}

    How have you gotten from the denominator in the first fraction ( \tan2x) to the denominator in the second fraction ( 1-\tan^2x)?


    p.s sorry for being frustrating, and thanks for your time.
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    (Original post by marcsaccount)
    ...but I don't understand the step you've taken here




    How have you gotten from the denominator in the first fraction ( \tan2x) to the denominator in the second fraction ( 1-\tan^2x)?


    p.s sorry for being frustrating, and thanks for your time.
    It's fine don't worry.

    Well you know our identity \tan(2A)=\dfrac{2\tan(A)}{1-\tan^{2}(A)}

    we let A=x and that gives us \tan(2x)=\dfrac{2\tan(x)}{1-\tan^{2}(x)}

    To get that on the denominator, instead of writing \tan(2x) we write \dfrac{2\tan(x)}{1-\tan^{2}(x)} on the denominator, hence the result above
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    (Original post by Robbie242)
    It's fine don't worry.

    Well you know our identity \tan(2A)=\dfrac{2\tan(A)}{1-\tan^{2}(A)}

    we let A=x and that gives us \tan(2x)=\dfrac{2\tan(x)}{1-\tan^{2}(x)}

    To get that on the denominator, instead of writing \tan(2x) we write \dfrac{2\tan(x)}{1-\tan^{2}(x)} on the denominator, hence the result above
    Ok, thanks for your help
 
 
 
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