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TransendentialPI
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I can't see the substitution I'm supposed to make. Any hints? See next for the exercise.
 one year ago
 one year ago
TransendentialPI Group Title
I can't see the substitution I'm supposed to make. Any hints? See next for the exercise.
 one year ago
 one year ago

This Question is Closed

cwrw238 Group TitleBest ResponseYou've already chosen the best response.0
try u = sqrt(2x)
 one year ago

TransendentialPI Group TitleBest ResponseYou've already chosen the best response.0
I 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.
 one year ago

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

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

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Now use Integration By parts..
 one year ago

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

waterineyes Group TitleBest 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..
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
you 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: \[\int\limits \sqrt{a^2  x^2}dx = \frac{1}{2}x \cdot \sqrt{a^2  x^2} + \frac{1}{2}a^2 \cdot \sin^{1} (\frac{x}{a}) + C\] \[\int\limits \frac{1}{\sqrt{a^2  u^2}}dx = \sin^{1}(\frac{x}{a}) + C\]
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Sorry  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: \[\int\limits\limits u \cdot \sin^{1}(u) du = \sin^{1}(u) \cdot \frac{u^2}{2} + \int\limits\limits( \frac{1}{\sqrt{1  u^2}} \times \frac{u^2}{2}) du\]
 one year ago

TransendentialPI Group TitleBest ResponseYou've already chosen the best response.0
Thanks, 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!
 one year ago

waterineyes Group TitleBest ResponseYou've already chosen the best response.1
Welcome dear..
 one year ago
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