trig integration (1 Viewer)

with-chu · May 10, 2010

Find, USING REVERSE CHAIN RULE,
integral of 1/[(sqrt x)(1+x)]

using 3u integration...

please helppp

Drongoski · May 10, 2010

$\int \ \frac{1}{\sqrt{x}(1+x)}\ dx\ \ =\ \ 2\int \frac{1}{1+x}\cdot \frac{1} {2\sqrt{x}} \ dx\\ \\ =\ \ 2\int \frac{1}{1+(\sqrt{x})^{2}}\ d\sqrt{x}\ \ =\ \ 2\ tan^{-1}\sqrt{x}\ +\ C$

This is really doing it the reversed chain rule way but few will see it that way.

Drongoski · May 10, 2010

Or by the conventional substitution method (which is a reversed chain rule method)

$$ Let u $ = \sqrt{x} \ \ \ \ \ \ \therefore \ \ \frac{\mathrm{d} u}{\mathrm{d} x}\ =\ \frac{1}{2\sqrt{x}}\ \ \ \therefore \ \ du\ =\ \frac{1}{2\sqrt{x}}dx\\ \\ \therefore \ \ \int \frac{1}{\sqrt{x}(1+x)}dx\ =\ 2\int \frac{1}{1+u^{2}}\ du\ \ =\ \ 2\ tan^{-1}u\ +\ C\\ \\ =\ \ 2\ tan^{-1}\sqrt{x}\ +\ C$

with-chu · May 11, 2010

I don't get the first method... if you take out one on root x from the denominator, why is it on the numerator?

I get the second method though. Is it likely they will specifically ask for the reverse chain rule to integrate something like that? They normally say 'using ... method, OR OTHERWISE', right?

Thanks for your help again! much appreciated.

Drongoski · May 11, 2010

with-chu said:
I don't get the first method... if you take out one on root x from the denominator, why is it on the numerator?

I get the second method though. Is it likely they will specifically ask for the reverse chain rule to integrate something like that? They normally say 'using ... method, OR OTHERWISE', right?

Thanks for your help again! much appreciated.

Oh . . I made a mistake. root x should be denominator. I'll correct it now. Thanks for alerting me to it.

trig integration (1 Viewer)

with-chu

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Drongoski

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Drongoski

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with-chu

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Drongoski

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