A community for students.
Here's the question you clicked on:
 0 viewing
eistein+newton+gallileo=_me
 4 years ago
find the integral of 1/root(x^24)
eistein+newton+gallileo=_me
 4 years ago
find the integral of 1/root(x^24)

This Question is Closed

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1315245910188:dw

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1let \[\sec(\theta)=\frac{x}{2}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[2 \sec(\theta)=x => 2\sec(\theta)\tan(\theta) d \theta=dx\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{}^{}\frac{1}{\sqrt{(2\sec(\theta))^24}} 2\sec(\theta) \tan(\theta) d \theta \]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\frac{2}{\sqrt{4}}\int\limits_{}^{}\frac{\sec(\theta) \tan(\theta)}{\sqrt{\sec^2(\theta)1}} d \theta \]

eistein+newton+gallileo=_me
 4 years ago
Best ResponseYou've already chosen the best response.0mmmmm the answer is ln[x + root(X^2 4) ] + c

eistein+newton+gallileo=_me
 4 years ago
Best ResponseYou've already chosen the best response.0mmmmm the answer is ln[x + root(X^2 4) ] + c .......negative 4

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{}^{}\frac{\sec(\theta) \tan(\theta)}{\sqrt{\tan^2(\theta)}} d \theta\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits_{}^{}\sec(\theta) d \theta =\int\limits_{}^{}\sec(\theta) \cdot \frac{\sec(\theta)+\tan(\theta)}{\sec(\theta)+\tan(\theta)} d \theta\] let \[u=\sec(\theta)+\tan(\theta)=>du=[\sec(\theta)\tan(\theta)+\sec^2(\theta)] d \theta\] so we have \[\int\limits_{}^{}\frac{du}{u}=\lnu+C\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1but \[u=\sec(\theta)+\tan(\theta)\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1so we have \[\ln\sec(\theta)+\tan(\theta)+C\] but we made the substition that \[\sec(\theta)=\frac{x}{2}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1so if \[\sec(\theta)=\frac{x}{2}\] dw:1315246444736:dw so \[\tan(\theta)=\frac{\sqrt{x^24}}{2}\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1so we actually have \[\ln\frac{x}{2}+\frac{\sqrt{x^24}}{2}+C\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1\[\ln\frac{x+\sqrt{x^24}}{2}+C=\lnx+\sqrt{x^24}\ln2+C\]

myininaya
 4 years ago
Best ResponseYou've already chosen the best response.1but ln(2) is just a constant so it can all be included in the C part so we could write \[\lnx+\sqrt{x^24}+K\]
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.