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anonymous

  • 4 years ago

Evaluate the integral (tanx+sinx)/secx dx from pi/6 to pi/3

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  1. anonymous
    • 4 years ago
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    \[\int\limits_{\pi/6}^{\pi/3} (tanx +sinx)/secx dx\]

  2. anonymous
    • 4 years ago
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    why not rewrite as integral of sin(x) + (1/2)sin(2x) ?

  3. campbell_st
    • 4 years ago
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    simplify the expression 1st \[\tan(x)/\sec(x) + \sin(x)/\sec(x) = sinx/\cos^2(x) + \tan(x)\]

  4. anonymous
    • 4 years ago
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    i got to the point where sinx + sinxcosx dx

  5. anonymous
    • 4 years ago
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    integral of sin(x)cos(x) is basically integral of sin(2x), although u-substituting u=sin(x) is also easy

  6. anonymous
    • 4 years ago
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    split and do the integral of the sin(x) separately from the sin(x)cos(x)

  7. campbell_st
    • 4 years ago
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    so the problem becomes \[\int\limits_{0}^{\pi/3} \sin(x)/\cos^2(x) dx + \int\limits_{0}^{\pi/3} \tan(x) \]

  8. anonymous
    • 4 years ago
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    i let u = cosx

  9. anonymous
    • 4 years ago
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    so i get -cosx -cos^2x/2 +c

  10. anonymous
    • 4 years ago
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    @campbell_st check your definition of 1/sec(x)

  11. campbell_st
    • 4 years ago
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    sec = 1/cos

  12. anonymous
    • 4 years ago
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    right, so tan(x)/sec(x) = ?

  13. anonymous
    • 4 years ago
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    sinx

  14. campbell_st
    • 4 years ago
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    and tan = sin/cos to tan/sec = sin/cos x 1/cos or sin/cos^2

  15. anonymous
    • 4 years ago
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    @Lammy -- you typed into OpenStudy "Evaluate the integral (tanx+sinx)/secx dx from pi/6 to pi/3" I cutpasted it into www.wolframalpha.com "Evaluate the integral (tanx+sinx)/secx dx from pi/6 to pi/3" and got 1/4 (-1+2 sqrt(3))

  16. anonymous
    • 4 years ago
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    @campbell_st -- no that should be tan/sec = (sin/cos) / (1/cos) or sin

  17. campbell_st
    • 4 years ago
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    lol... sorry about the boundary mis-typing...

  18. campbell_st
    • 4 years ago
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    for the 1st integral let u = cos x then du = - sin(x) dx so the integral is \[\int\limits_{?}^{} du/u^2\]

  19. campbell_st
    • 4 years ago
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    you'll need to find the boundary values...

  20. anonymous
    • 4 years ago
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    the boundary is 1/2 and sqrt( 3)/2

  21. campbell_st
    • 4 years ago
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    oops it should be \[- \int\limits_{\sqrt{3}/2}^{1/2} du/u^2 \] the other part is \[\int\limits_{\pi/6}^{\pi/3} \tan(x) dx\] so

  22. anonymous
    • 4 years ago
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    ok i think i figure it out

  23. anonymous
    • 4 years ago
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    i got sqrt3/2 - 1/4 same as wolfram

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