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Integrate [1+tanx.tan(x+a)]dx

  1. Dec 16, 2012 #1
    1. The problem statement, all variables and given/known data
    ∫[(1+tanx.tan(x+a)]dx


    2. Relevant equations



    3. The attempt at a solution
    ∫sec^x.tan(x+a)
    after that i don't know as i tried method of substitution by putting x+a=t but i am not getting the answer as its form on the whole remains same
     
  2. jcsd
  3. Dec 16, 2012 #2

    sharks

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    Your problem isn't clear. It is missing closing parentheses.

    Is this what you meant?
    $$\int (1+ \tan x).\tan (x+a).dx$$
    or this?
    $$\int (1+ \tan x.\tan (x+a)).dx$$
     
  4. Dec 16, 2012 #3

    HallsofIvy

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    As sharks said, what you wrote is ambiguous- you have a "(" without a corresponding ")". Also 1+ tan x is NOT sec^2 x. Did you mean 1+ tan^2 x.
     
  5. Dec 16, 2012 #4

    lurflurf

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    ∫[(1+tanx)tan(x+a)]dx
    is a usual calculus problem the other form is not.
     
  6. Dec 16, 2012 #5

    Curious3141

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    Hint: angle sum formula for tangent. Simplify with trig identity before integration.

    EDIT: assuming the problem is: $$\int (1+ \tan x.\tan (x+a)).dx$$
     
  7. Dec 16, 2012 #6
    apologies for my mistake it is ∫ [(1+tanx)tan(x+a)]dx
     
  8. Dec 16, 2012 #7

    Curious3141

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    That can also be solved the same way. At least, that's the way I would do it. Expand out the brackets, then use my hint on the second term.
     
  9. Dec 16, 2012 #8
    i am applying that but its getting complex,what i think is we have to multiply and divide the equation by some constant entity like cot a or anything else as i"ll show you a example
    integrate 1/sin(x-a)cos(x-b)dx
    in order to solve it we divide and multiply it by cos(a-b) then we"ll expand numerator as cos{(x-b)-(x-a)} and then apply c-d formula
     
  10. Dec 16, 2012 #9

    Curious3141

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    Why don't you show what you did? Applied to this problem, not some other. And please use latex formatting.
     
  11. Dec 16, 2012 #10

    sharks

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    First, expand your expression, step by step. As Curious3141 said, show us your work. Try to use LaTeX to make your expressions more readable. See the link below my reply, in my forum signature.

    $$\int (1+ \tan x).\tan (x+a).dx=\int \tan (x+a).dx + \int \tan x.\tan (x+a).dx$$
    The first integral should be easy. If not, you won't be able to do the second one.

    To find the second integral, here's a hint. Just expand ##\tan (x+a)## and re-arrange the terms so you get ##\tan x.\tan (x+a)## on the L.H.S.
     
  12. Dec 16, 2012 #11
    I am just new on PF and i am just learning to use LATEX formatting and my approach is
    (1+tanx)tan(x+a)
    =>(1+tanx){tanx+tana/1-(tan x)(tana)}
    =>(1+tanx){1+(sinx/cos)/1-(sinx/cosx)(sina/cosa),then taking L.C.M and converting 1+tanx to (cosx+sinx)/cosx
    =>{(cosx+sinx)/cosx}(cosx+sinx)/cosx(cosx.cosa)/(1-sinxsina)
    then in the end its coming out to be cosa∫1/{cosx(1-sinxsina)}
    and i don't know how to solve this
     
    Last edited: Dec 16, 2012
  13. Dec 16, 2012 #12
    The method in 10 will get you the answer easily.
    Manipulate the expression for tan(x+a) to find (tan x)(tan[x+a])
     
  14. Dec 16, 2012 #13
    i am not getting the answer ,the answer is cot a.log{sec(x+a)\secx}+c
     
  15. Dec 16, 2012 #14
    I am getting that as one of the terms if log==natural log and {} are absolute value but it is not the complete solution.
     
  16. Dec 16, 2012 #15

    sharks

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    There is no need to expand tan into sin and cos. Re-read my previous post #10.

    The answer that you've stated is not complete, as it deals with the second integral only.

    Can you integrate the following?
    $$\int \tan (x+a).dx$$

    You should get:
    $$\ln |\sec (x+a)|$$
     
  17. Dec 16, 2012 #16

    SammyS

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    What answer do you get ?

    Remember, [itex]\displaystyle \log(\sec(x+a)/\sec(x))=\log(\sec(x+a)\cos(x))=\log(\sec(x+a))+ \log(\cos(x))\ .[/itex]
     
  18. Dec 16, 2012 #17

    Curious3141

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    Note that:

    $$\tan (A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$$

    Now put:

    $$A = x + a, B = x$$

    and see what you get.
     
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