A community for students.
Here's the question you clicked on:
 0 viewing
Jonask
 2 years ago
prove intergral
Jonask
 2 years ago
prove intergral

This Question is Closed

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0\[\int\limits _0^1\frac{1}{\alpha\cos x}dx=\frac{\pi}{\sqrt{\alpha^21}}\]

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0i tried \[\tan \frac{x}{2}=u\] i have \[\huge \int\limits \frac{1}{\alpha\frac{1t^2}{1+t^2}}\frac{2dt}{1+t^2}=2\huge \int\limits \frac{1}{\alpha1+(1+\alpha)t^2}dt\]

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0\[\huge \frac{2}{\alpha+1}\int\limits \frac{dt}{(\sqrt{\frac{\alpha1}{\alpha+1}})^2+t^2}=\frac{2}{\alpha+1}\tan^{1}(\frac{t}{\sqrt{\frac{\alpha1}{\alpha+1}}})\]

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0first of all tan pi/2 is undefined,is there anything wrong here?

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0oh something missing the whole thing shud be divided by \[\sqrt{\frac{\alpha1}{\alpha+1}}\]

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0\[\huge \frac{2}{\sqrt{\alpha^21}}\tan^{1}\frac{\frac{x}{2}}{\sqrt{\frac{\alpha1}{\alpha+1}}}\]

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.0Did you mutliply by conjugate?

linshan789
 2 years ago
Best ResponseYou've already chosen the best response.0Wait, you're proving. My bad, I thought you were solving using trig sub.

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0same thing if you can solve it then you actually proved it

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0my problem here is tan pi/2 is undefined

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0METHOD II RATIONALISE AND SEPERATE \[\int\limits \frac{1}{\alpha\cos x}dx=\int\limits \frac{\alpha+\cos x}{\alpha^2\cos^2x}dx=\int\limits \frac{\alpha}{\alpha^2\cos^2x}+\int\limits \frac{\cos x}{\alpha^2\cos x}\]

linshan789
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1363445220695:dw

linshan789
 2 years ago
Best ResponseYou've already chosen the best response.0Did you get anything like that?

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1363452681715:dw i thought thig is +

linshan789
 2 years ago
Best ResponseYou've already chosen the best response.0Did you get anything like that?

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0yes thats my first solution there on top instead of z i used t

phi
 2 years ago
Best ResponseYou've already chosen the best response.0I got \[ \int\limits _0^1\frac{1}{\alpha\cos x}dx= \frac{2}{\sqrt{\alpha^21}} \tan^{1}\left(\sqrt{\frac{\alpha+1}{\alpha1}}\tan\frac{1}{2}\right)\] I don't see the inverse tan giving pi/2 except at alpha>1

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0can you explain how you did it

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0these makes it clear ,thanks but how do we simplify to get the final answer

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{\pi}{\sqrt{\alpha^21}}\]

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0initially \[\int _0^\pi\]

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0i had typed the wrong boundaries 0 to 1 instead of 0 to pi

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0SO THE PROBLEM IS HERE \[\frac{1}{\sqrt{\alpha^21}}\tan^{1}\left( {\sqrt{\frac{\alpha1}{\alpha+1}}\tan (x/2)}\right)\]

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{\sqrt{\alpha^21}}\tan^{1}\left( {\sqrt{\frac{\alpha1}{\alpha+1}}\tan (\pi/2)}\right)\] the part inside is undefined

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0\[\frac{2}{\sqrt{\alpha^21}}\tan^{1}\left( {\sqrt{\frac{\alpha1}{\alpha+1}}\tan (\pi/2)}\right)\] i mean 2

phi
 2 years ago
Best ResponseYou've already chosen the best response.0that makes more sense, because I was going to say the original expression is false. however, for this new limit, you can say tan(x) as x> pi/2 approaches infinity and the inverse tangent of x> infinity approaches pi/2

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0aha eureka!! thanks guys

phi
 2 years ago
Best ResponseYou've already chosen the best response.0Here is the problem using the correct limits

Jonask
 2 years ago
Best ResponseYou've already chosen the best response.0thanks again for helping sry for the error
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.