25-26 Find the length of the arc of the curve from point P to point Q. 25. y = 1/2 x^2, P(-1, 1/2), Q(1, 1/2) sqrt(2) + ln(1+sqrt(2)) y' = x integral sqrt(1+x^2) integral from -1 to 1 sqrt(1+x^2) dx tan theta = x dx = sec^2 theta d theta integral from -1 to 1 sqrt(1+tan^2 theta) * sec^2 theta d theta integral from -1 to 1 sec theta * sec^2 theta d theta integral u v' = uv - integral u' v u = sec theta, v = tan theta du = sec theta tan theta, dv = sec^2 theta sec theta tan theta - integral tan theta * sec theta tan theta d theta sec theta tan theta - integral tan^2 theta sec theta d theta
Added by Joseph H.
Close
Step 1
Let $u = \sec(\theta)$ and $dv = \tan(\theta) \, d\theta$. Then, $du = \sec(\theta) \tan(\theta) \, d\theta$ and $v = \int \tan(\theta) \, d\theta = -\ln|\cos(\theta)|$. ** Show more…
Show all steps
Your feedback will help us improve your experience
Gopesh Vishwakarma and 92 other Calculus 2 / BC educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
I'm not sure if I am doing this correctly. Is it suppose to be integration by parts? Also the integration by parts seems like it will go on forever since both parts are trig functions. Can you please explain showing all work and not just plugging into some formula in the back of the book? Thank you.
Suzanne W.
I got stuck on the integration by parts. I'm not sure if it is even possible because the two parts are both trig functions and it seems like it will go on forever. This is assuming I've even been doing the arc length problem correctly. Can you please show me how to do this problem showing all work? Please don't just use a formula from the back of the book and plug things in. I need to show all work and plus I need to learn how to do these integration by parts with trig functions properly. Thank you.
Gopesh V.
Can someone explain how to solve this please? Also explain what to do when it is integration by parts but with trig functions? It seems like the integration will go on forever if it is using trig functions.
Recommended Textbooks
Calculus: Early Transcendentals
Thomas Calculus
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD