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

This Question is Closed

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

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

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

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

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