How to do tan inverse integraL? (1 Viewer)

malcolm21 · Aug 6, 2015

Find f(x)

InteGrand · Aug 6, 2015

malcolm21 said:
Find f(x)

$f^\prime (x) = \frac{1}{4+9x^2}$

$\Rightarrow f(x) = \int \frac{1}{4+9x^2}\text{ d}x$

$$= \frac{1}{9}\int \frac{1}{\frac{4}{9}+x^2}\text{ d}x$ (taking out a factor of $9$ from the denominator)$

$$= \frac{1}{9}\times \sqrt{\frac{9}{4}}\tan^{-1} \left(\sqrt{ \frac{9}{4}}x \right)+c$ (standard integral)$

$= \frac{1}{9}\times \frac{3}{2}\tan^{-1} \left( \frac{3}{2}\cdot x \right)+c$

$$=\frac{1}{6}\tan^{-1} \left( \frac{3x}{2} \right)+c$.$

kawaiipotato · Aug 6, 2015

$$ If your function is in the form of $ \frac{1}{a^2 + b^{2}x^{2}}$

$$ Then $ \int \frac{1}{a^2 + b^{2}x^{2}} dx = \frac{1}{ab}tan^{-1}(\frac{bx}{a})$
$tan inverse $ ( $ square root of $ b^2 x^2 $ divide by square root of a^2 $ ) $) all over 'a' and number in front of x 'b'$

$$ Hence, $ \int \frac{1}{4+9x^{2}} dx$

$= \frac{1}{3*2}tan^{-1}(\frac{3x}{2}) + $ C $$

edit: oops integrand already posted

malcolm21 · Aug 6, 2015

kawaiipotato said:
$$ If your function is in the form of $ \frac{1}{a^2 + b^{2}x^{2}}$

$$ Then $ \int \frac{1}{a^2 + b^{2}x^{2}} dx = \frac{1}{ab}tan^{-1}(\frac{bx}{a})$
$tan inverse $ ( $ square root of $ b^2 x^2 $ divide by square root of a^2 $ ) $) all over 'a' and number in front of x 'b'$

$$ Hence, $ \int \frac{1}{4+9x^{2}} dx$

$= \frac{1}{3*2}tan^{-1}(\frac{3x}{2}) + $ C $$

edit: oops integrand already posted

Does sin inverse have something similar? Never seen that 1/ab thing, but it works!

kawaiipotato · Aug 6, 2015

malcolm21 said:
Does sin inverse have something similar? Never seen that 1/ab thing, but it works!

same thing except you don't divide by 'a' which is the constant term under the sqrt so:

$\int \frac{1}{\sqrt{a^2 - b^{2}x^{2}}} dx$

$= \frac{1}{b} sin^{-1}{\frac{bx}{a}} + $ C $$
always divide by number in front of x

malcolm21 · Aug 7, 2015

kawaiipotato said:
same thing except you don't divide by 'a' which is the constant term under the sqrt so:

$\int \frac{1}{\sqrt{a^2 - b^{2}x^{2}}} dx$

$= \frac{1}{b} sin^{-1}{\frac{bx}{a}} + $ C $$
always divide by number in front of x

not sure if you're still on but do you know if there's a pattern like this when findignt eh derivative of the inverse functions?

InteGrand · Aug 7, 2015

malcolm21 said:
not sure if you're still on but do you know if there's a pattern like this when findignt eh derivative of the inverse functions?

Are you after these?:

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\sin ^{-1} (ax+b) \right)=\frac{a}{\sqrt{1-(ax+b)^2}}$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\cos ^{-1} (ax+b) \right)=\frac{-a}{\sqrt{1-(ax+b)^2}}$

$\frac{\mathrm{d}}{\mathrm{d}x}\left(\tan ^{-1} (ax+b) \right)=\frac{a}{1+(ax+b)^2}$

(Just comes from applying the chain rule)

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How to do tan inverse integraL? (1 Viewer)

malcolm21

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InteGrand

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kawaiipotato

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malcolm21

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kawaiipotato

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malcolm21

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InteGrand

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