anonymous
  • anonymous
Evaluate the integral (tanx+sinx)/secx dx from pi/6 to pi/3
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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 \[\tan(x)/\sec(x) + \sin(x)/\sec(x) = sinx/\cos^2(x) + \tan(x)\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.