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So I am working on an integral part of it involves $$\int (3t^2-4)^\frac{5}{2}$$

Obviously the sub is $$\Biggl[3\Biggl(\frac{2}{\sqrt{3}}sec(t)\Biggl)^2-4\Biggl]^{\frac{5}{2}}$$

which becomes

$$2\sqrt{tan^2(t)}^5$$ But how is that possible the $5$ is supposed to stay with whats inside the square root I fail to see where you can just pull out the 5 from inside the radical and say that its actually the radical and everything inside to the 5th power and not just what is inside the radical to the fifth power.

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  • $\begingroup$ It should be $$32\left(\sqrt{\tan^2 x}\right)^5$$ $32,$ not $2.$ $\endgroup$ Aug 27 '19 at 22:18
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The property of exponents that matters is $\left(a^b\right)^c=a^{bc}=\left(a^c\right)^b$ for $a \gt 0$. If you have a square root sign, $b$ or $c$ is $\frac 12$ and this still applies. In fact, when you write $(3t^2-4)^{\frac 52}$ you are not indicating whether this is $\sqrt{(3t^2-4)^5}$ or $\left(\sqrt{3t^2-4}\right)^5$. It doesn't matter because they are equal by the preceding property.

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    $\begingroup$ I just edited to fix a small typo $\endgroup$
    – peek-a-boo
    Aug 27 '19 at 20:55

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