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TransendentialPI
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This exercise is in a section about using trig substitution to integrate. I can use trig sub, but I'm just not seeing what initial substitution to use so I can apply trig sub. Can anyone give me a hint? I don't need (don't want) it solve to the end. At least not yet.
 2 years ago
 2 years ago
TransendentialPI Group Title
This exercise is in a section about using trig substitution to integrate. I can use trig sub, but I'm just not seeing what initial substitution to use so I can apply trig sub. Can anyone give me a hint? I don't need (don't want) it solve to the end. At least not yet.
 2 years ago
 2 years ago

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TransendentialPI Group TitleBest ResponseYou've already chosen the best response.0
See attached
 2 years ago

Davidjohn Group TitleBest ResponseYou've already chosen the best response.2
dw:1342935015183:dw
 2 years ago

Davidjohn Group TitleBest ResponseYou've already chosen the best response.2
woops cos theta/4 = 1/ hypotenuse, not just the hypotenuse
 2 years ago

Davidjohn Group TitleBest ResponseYou've already chosen the best response.2
just a guess with what is out there
 2 years ago

Davidjohn Group TitleBest ResponseYou've already chosen the best response.2
dw:1342935652408:dw actually this looks better
 2 years ago

Davidjohn Group TitleBest ResponseYou've already chosen the best response.2
then dw:1342935809458:dw
 2 years ago

dumbcow Group TitleBest ResponseYou've already chosen the best response.0
hint: use trig identity \[\sec^{2} = 1+\tan^{2}\] Factor out the 16 from radical in denominator \[\sqrt{16+x^{6}} = \sqrt{16}\sqrt{1+\frac{x^{6}}{16}}\] Now make the substitution \[x^{6} = 16\tan^{2}\] hope that helps
 one year ago
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