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Really quick question. Where did the 1/81 come from..
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Ah. what is your x-value? the thing you used to sub.
\[x=3\sin (\theta)\]
\[\large \int\limits\frac{1}{x^4\sqrt{9-x^2}}dx\]\[\large \sqrt{9-x^2} : x= 3\sin(\theta)\]\[\large dx = 3\cos(\theta)d(\theta)\]
ok, so if you have your value of x.. then \(\large (3\sin(\theta))^4\) = \(\large 81\sin^4(\theta)\)
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lol smh
there you go! :)
wow... I need to get to bed. Thanks lol
Hahaha no problem.
And thank you :D
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:)
you're welcome, Ms. Jhannybean
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