anonymous
  • anonymous
Evaluate the integral (tanx+sinx)/secx dx from pi/6 to pi/3
Mathematics
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anonymous
  • anonymous
Evaluate the integral (tanx+sinx)/secx dx from pi/6 to pi/3
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\int\limits_{\pi/6}^{\pi/3} (tanx +sinx)/secx dx\]
anonymous
  • anonymous
why not rewrite as integral of sin(x) + (1/2)sin(2x) ?
campbell_st
  • campbell_st
simplify the expression 1st \[\tan(x)/\sec(x) + \sin(x)/\sec(x) = sinx/\cos^2(x) + \tan(x)\]

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anonymous
  • anonymous
i got to the point where sinx + sinxcosx dx
anonymous
  • anonymous
integral of sin(x)cos(x) is basically integral of sin(2x), although u-substituting u=sin(x) is also easy
anonymous
  • anonymous
split and do the integral of the sin(x) separately from the sin(x)cos(x)
campbell_st
  • campbell_st
so the problem becomes \[\int\limits_{0}^{\pi/3} \sin(x)/\cos^2(x) dx + \int\limits_{0}^{\pi/3} \tan(x) \]
anonymous
  • anonymous
i let u = cosx
anonymous
  • anonymous
so i get -cosx -cos^2x/2 +c
anonymous
  • anonymous
@campbell_st check your definition of 1/sec(x)
campbell_st
  • campbell_st
sec = 1/cos
anonymous
  • anonymous
right, so tan(x)/sec(x) = ?
anonymous
  • anonymous
sinx
campbell_st
  • campbell_st
and tan = sin/cos to tan/sec = sin/cos x 1/cos or sin/cos^2
anonymous
  • anonymous
@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))
anonymous
  • anonymous
@campbell_st -- no that should be tan/sec = (sin/cos) / (1/cos) or sin
campbell_st
  • campbell_st
lol... sorry about the boundary mis-typing...
campbell_st
  • campbell_st
for the 1st integral let u = cos x then du = - sin(x) dx so the integral is \[\int\limits_{?}^{} du/u^2\]
campbell_st
  • campbell_st
you'll need to find the boundary values...
anonymous
  • anonymous
the boundary is 1/2 and sqrt( 3)/2
campbell_st
  • campbell_st
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
anonymous
  • anonymous
ok i think i figure it out
anonymous
  • anonymous
i got sqrt3/2 - 1/4 same as wolfram

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