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anonymous
 one year ago
ques
anonymous
 one year ago
ques

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0How may I solve the following integral? \[\int\limits (16x^2)^{\frac{3}{2}}dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I know for \[\int\limits (a^2x^2)^\frac{1}{2}dx=\frac{x}{2}.(a^2x^2)^\frac{1}{2}+\frac{a^2}{2}\sin^{1}(\frac{x}{a})\] But this is cubed now

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1x = 4 sint dx = 4 cost dt x^2 = 16 sin^2 t 16 x^2 = 16 cos^2 t

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1(16 x^2)^3/2 = 4^3 cos^3t ok?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[I=256 \int\limits \cos^4(t)dt\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thanks a lot, let me try it now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok so I'm getting a really weird answer \[I=96 \sin^{1}({\frac{x}{4}})+8x\sqrt{16x^2}+\frac{x\sqrt{16x^2}(8x^2)}{32}\] Now I can't say for sure how much accurate this answer is, but more or less I now understand how to do that integral thanks @Loser66
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