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Math Help - Solve The Following Trigonometric Integral.

  1. #1
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    Solve The Following Trigonometric Integral. (1)

    Hi
    i tried to do it many times
    but really its hard
    its from an incomplete exam

    \int \frac{1-tan(x)}{1+tan(x)}dx
    Last edited by TWiX; January 18th 2010 at 07:46 AM. Reason: tittle
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  2. #2
    Super Member General's Avatar
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    Quote Originally Posted by TWiX View Post
    Hi
    i tried to do it many times
    but really its hard
    its from an incomplete exam

    \int \frac{1-tan(x)}{1+tan(x)}dx
    Hello .
    Multiply the integrated function by \frac{1-tan(x)}{1-tan(x)}
    You will get:

    \int\frac{(1-tan(x))^2}{1-tan^2(x)}dx
    = \int\frac{tan^2(x)-2tan(x)+1}{1-tan^2(x)}dx
    =\int\frac{sec^2(x)-2tan(x)}{1-tan^2(x)}dx
    Multiply the integrated function by \frac{cos^2(x)}{cos^2(x)}<br />
, And with some trigonometric identities, you will face:
    \int\frac{1-sin(2x)}{cos(2x)}dx

    Now, It looks easier. Right ?
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  3. #3
    MHF Contributor Calculus26's Avatar
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    See attachment
    Attached Thumbnails Attached Thumbnails Solve The Following Trigonometric Integral.-trigint.jpg  
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  4. #4
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    Hello, =TWiX!

    \int \frac{1-\tan x}{1+\tan x}\,dx
    The function is: . \frac{1-\tan x}{1 + \tan x} \;=\;\frac{1-\frac{\sin x}{\cos x}}{1 + \frac{\sin x}{\cos x}}


    Multiply by \frac{\cos x}{\cos x}\!:\quad\frac{\cos x\left(1 - \frac{\sin x}{\cos x}\right)}{\cos x\left(1 + \frac{\sin x}{\cos x}\right)} \;=\; \frac {\cos x - \sin x}{\cos x + \sin x}


    The integral becomes: . \int\frac{\cos x - \sin x}{\cos x + \sin x}\,dx


    Let: . u \:=\:\cos x + \sin x \quad\Rightarrow\quad du \:=\:(-\sin x + \cos x)\,dx \quad\Rightarrow\quad du \:=\:(\cos x - \sin x)\,dx


    Substitute: . \int\frac{du}{u}\quad\hdots Got it?

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  5. #5
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    Thanks all.
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