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anonymous
 4 years ago
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.
anonymous
 4 years ago
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.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342935015183:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0woops cos theta/4 = 1/ hypotenuse, not just the hypotenuse

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just a guess with what is out there

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342935652408:dw actually this looks better

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then dw:1342935809458:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hint: 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
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