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siddarth95

  • 3 years ago

Help prove this integral

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  1. siddarth95
    • 3 years ago
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    \[\int\limits_{0}^{\pi/2}\frac{ dx }{ \sin x + \cos x } = \frac{ 1 }{ \sqrt{2} } \ln \ \frac{ \sqrt{2}+1 }{ \sqrt{2}-1 }\]

  2. Jonask
    • 3 years ago
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    let u=sin x +cos x du=cosx-sinx dx

  3. Jonask
    • 3 years ago
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    \[\int \frac{1}{ \cos x+\sin x}dx\] multiply by cos x-sin x both numerator and denominator

  4. Jonask
    • 3 years ago
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    \[\int \frac{\cos x-\sin x}{\cos^2x-\sin^2x} dx=\int \frac{\cancel{\cos x-\sin x}du}{\cos 2x}\frac{du}{\cancel{\cos x-\sin x}}\]

  5. Jonask
    • 3 years ago
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    \[\int \frac{du}{\cos 2x}=\int \sec 2x=\ln|\sec 2x+\tan 2x|\]/2

  6. Jonask
    • 3 years ago
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    error i never substituted u

  7. Meepi
    • 3 years ago
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    I don't have time to put an explanation here, but use the substitution u = tan(x/2): http://www-math.mit.edu/~djk/18_01/chapter24/section03.html

  8. Meepi
    • 3 years ago
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    Pretty sure others can help you if you get stuck :3

  9. experimentX
    • 3 years ago
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    |dw:1362839531405:dw|

  10. siddarth95
    • 3 years ago
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    its all good .. thanks for the responses :)

  11. experimentX
    • 3 years ago
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    there's one another way to do it .. change all trigs into half angles, and change sines and cosines into tan and sec .. you should end up something like|dw:1362841238898:dw|

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