A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years ago
I can't see the substitution I'm supposed to make. Any hints? See next for the exercise.
 2 years ago
I can't see the substitution I'm supposed to make. Any hints? See next for the exercise.

This Question is Closed

TransendentialPI
 2 years ago
Best ResponseYou've already chosen the best response.0I did dw:1343307585147:dwdw:1343307655512:dw I'm supposed to be geting this into the form so I can find the formula in a integral table.

TransendentialPI
 2 years ago
Best ResponseYou've already chosen the best response.0I did find this one, but it is making the integral more complicated:

TransendentialPI
 2 years ago
Best ResponseYou've already chosen the best response.0That is not getting me to one of those MC answers/

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1Now use Integration By parts..

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1\[\frac{d}{dx}(\sin^{1}x) = \frac{1}{\sqrt{1  x^2}}\]

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1\[\int\limits u \cdot \sin^{1}(u) du = \sin^{1}(u) \cdot \frac{u^2}{2}  \int\limits( \frac{1}{\sqrt{1  u^2}} \times \frac{u^2}{2}) du\] Try to solve for last part..

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1you can do it like this: \[ \frac{1}{\sqrt{1  u^2}} \times u^2 = \frac{1  u^2 + 1}{\sqrt{1  u^2}} =\sqrt{1  u^2} + \frac{1}{\sqrt{1  u^2}}\] Use the following two Special Integral Formula: \[\int\limits \sqrt{a^2  x^2}dx = \frac{1}{2}x \cdot \sqrt{a^2  x^2} + \frac{1}{2}a^2 \cdot \sin^{1} (\frac{x}{a}) + C\] \[\int\limits \frac{1}{\sqrt{a^2  u^2}}dx = \sin^{1}(\frac{x}{a}) + C\]

waterineyes
 2 years ago
Best ResponseYou've already chosen the best response.1Sorry  will be there in between terms: \[\frac{1}{\sqrt{1  u^2}} \times u^2 = \frac{1  u^2  1}{\sqrt{1  u^2}} =\sqrt{1  u^2}  \frac{1}{\sqrt{1  u^2}}\] By this your integration middle term will become positive: \[\int\limits\limits u \cdot \sin^{1}(u) du = \sin^{1}(u) \cdot \frac{u^2}{2} + \int\limits\limits( \frac{1}{\sqrt{1  u^2}} \times \frac{u^2}{2}) du\]

TransendentialPI
 2 years ago
Best ResponseYou've already chosen the best response.0Thanks, I just flipped back in the textbook section and they provide an example in the form \[\int\limits xsin^{1} x \] dx= that integral and then to use a the table again. Thank so much!
Ask your own question
Ask a QuestionFind 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.