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FP2 - Integrating derivatives of inverse trig funtions.

If y = arccos2x, then dy/dx = -2/sqrt(1-4x^2).

But surely integrating -2/sqrt(1-4x^2), would give 2arccos2x (+c), wouldn't it?

The issue I'm having here is that I can't see why the integral of 1/(a^2 + x^2) = 1/a (arctan(x/a)) yet for 1/sqrt(a^2-x^2) the answer is simply arcsin(x/a), with the 1/a omitted. It just seems to ignore the reverse chain rule altogether.

Can anyone help me with this? :o:
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Avatar for nuodai
nuodai
17
The best way to demonstrate this would be to make the substitution x=12sin⁡ux = \dfrac{1}{2}\sin u and integrate it that way. Same goes for the inverse tan.

If you have
Unparseable latex formula:

\displaystyle \int \dfrac{1}{\sqrt{a^2 - b^2x^2}}\ \mbox{d}x

, then you should use x=absin⁡ux = \dfrac{a}{b}\sin u.

If you have
Unparseable latex formula:

\displaystyle \int \dfrac{1}{a^2 + b^2x^2}\ \mbox{d}x

, then you should use x=abtan⁡ux = \dfrac{a}{b}\tan u.

You'll see what happens :p:

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