A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 years ago
Evaluate the integral (tanx+sinx)/secx dx from pi/6 to pi/3
anonymous
 4 years ago
Evaluate the integral (tanx+sinx)/secx dx from pi/6 to pi/3

This Question is Closed

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits_{\pi/6}^{\pi/3} (tanx +sinx)/secx dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0why not rewrite as integral of sin(x) + (1/2)sin(2x) ?

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0simplify the expression 1st \[\tan(x)/\sec(x) + \sin(x)/\sec(x) = sinx/\cos^2(x) + \tan(x)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got to the point where sinx + sinxcosx dx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0integral of sin(x)cos(x) is basically integral of sin(2x), although usubstituting u=sin(x) is also easy

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0split and do the integral of the sin(x) separately from the sin(x)cos(x)

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0so the problem becomes \[\int\limits_{0}^{\pi/3} \sin(x)/\cos^2(x) dx + \int\limits_{0}^{\pi/3} \tan(x) \]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i get cosx cos^2x/2 +c

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@campbell_st check your definition of 1/sec(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0right, so tan(x)/sec(x) = ?

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0and tan = sin/cos to tan/sec = sin/cos x 1/cos or sin/cos^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@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
 4 years ago
Best ResponseYou've already chosen the best response.0@campbell_st  no that should be tan/sec = (sin/cos) / (1/cos) or sin

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0lol... sorry about the boundary mistyping...

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0for the 1st integral let u = cos x then du =  sin(x) dx so the integral is \[\int\limits_{?}^{} du/u^2\]

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0you'll need to find the boundary values...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the boundary is 1/2 and sqrt( 3)/2

campbell_st
 4 years ago
Best ResponseYou've already chosen the best response.0oops 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
 4 years ago
Best ResponseYou've already chosen the best response.0ok i think i figure it out

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i got sqrt3/2  1/4 same as wolfram
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.