Question

calculus

Answer 1

Hmm, do you know the derivatives of inverse hyperbolic functions? that would help alot...

Answer 2

w8

Answer 3

Okay, but do you know the derivatives of them?

Answer 4

yes

Answer 5

http://math.info/image/64/deriv_hyp_archyp.gif

Answer 6

Okay, see how what we got looks a lot like arcsech?

Answer 7

do a \(u=3x\) sub and you'll basically have arcsech.

Answer 8

hmm let\[ u= \sqrt{1-9x^2}~~...(1) \\ \\ du=-\frac{9x}{\sqrt{1-9x^2}}dx...(2)\]
From (1),\[u= \sqrt{1-9x^2} \\ \\ \frac{u^2-1}{x}=-9x\]
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From (2),
\[du=(-9x)\frac{1}{\sqrt{1-9x^2}}dx\]
\[du=(\frac{u^2-1}{x})\frac{1}{\sqrt{1-9x^2}}dx\]
\[\frac{1}{u^2-1}du=\frac{1}{x\sqrt{1-9x^2}}dx\]
Then,
\[\int\limits \frac{1}{u^2-1}du\]

Answer 9

\[-\int\limits \frac{1}{1-u}du \\ \\ \text{since} \frac{d}{dx}(\tanh^{-1}(x))=\frac{1}{1-x^2} \\ \\ -\int\limits \frac{1}{1-u}du=-\tanh^{-1}u \\ \\ \Large =[-\tanh^{-1}(\sqrt{1-9x^2})]_{1/5}^{4/15}\]

Answer 10

As wio said, \[\large \frac{ d }{dx } arcsech x = \frac{ -1 }{x \sqrt{1 - x^2} }\]\[\large \int\limits_{}^{}\frac{ 1 }{ x \sqrt{1-(3x)^2} } dx\]
let u = 3x so x = u/3 and du/3 =dx:
\[\large \int\limits\limits\limits_{}^{}\frac{ 1 }{ \frac{ u }{ 3 } \sqrt{1-u^2} } \frac{ du }{ 3 } =\large \int\limits\limits_{}^{}\frac{ 1 }{ u \sqrt{1-u^2} } du\]
which looks a lot like the derivative of negative arcsech(u).