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Math Help - [SOLVED] Integrating trig functions

  1. #1
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    [SOLVED] Integrating trig functions

    \int{\sqrt{sec^2x}}

    The square root is giving me some problems here. I know that the identity of sec^2 x is tan x, but how does the sqrt play here?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kl.twilleger View Post
    \int{\sqrt{sec^2x}}
    the square root cancels the square. (technically you would have to use the absolute value, but it usually doesn't cause problems if we forget it, especially for indefinite integrals).

    so you want \int \sec x~dx

    now there's an interesting integral ...if you don't know it by heart, that is

    The square root is giving me some problems here. I know that the identity of sec^2 x is tan x, but how does the sqrt play here?
    the identity is \sec^2 x = 1 + \tan^2x. but you don't need that here
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  3. #3
    Super Member 11rdc11's Avatar
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    just integrate

    \int \sec{x}~dx

    To integrate \sec{x} multiply by \frac{\sec{x} + \tan{x}}{\sec{x} + \tan{x}}
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  4. #4
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    DUH!!!

    ok... so the integration is between 0 and pi/4 so it will be

    ln(sec\frac{\pi}{4} + tan\frac{\pi}{4}) - ln(sec 0 + tan 0) and then just plug and chug
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by kl.twilleger View Post
    DUH!!!

    ok... so the integration is between 0 and pi/4 so it will be

    ln(sec\frac{\pi}{4} + tan\frac{\pi}{4}) - ln(sec 0 + tan 0) and then just plug and chug
    yes
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