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I am in the tricky situation of trying to integrate the following.

$$\sqrt{4 a^2 (y-b)^2+c^4}$$

$a, b$ and $c$ are all known constants.

Can anybody provide insight as to how to do this?

I have tried to rearrange to fit the form:

$$\int (ax+b)^{\alpha}dx = \dfrac1a \cdot \dfrac{(ax+b)^{\alpha+1}}{\alpha+1} + \text{ constant}$$

But do not seem able to do so. Maybe excessive toiling has hidden an obvious answer from my eyes.

Thanks for the help.

Henry Walton
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4 Answers4

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$$\sqrt{4 a^2 (y-b)^2+c^4}=\dfrac1{2|a|}\sqrt{(y-b)^2+\left(\dfrac{c^2}{2a}\right)^2}$$

Using Trigonometric substitutions , set $y-b=\dfrac{c^2}{2a}\cdot\tan u$

and use How to integrate $\sec^3 x \, dx$?

Or Indefinite integral of secant cubed

Community
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lab bhattacharjee
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Let $$\displaystyle I = \int \sqrt{4a^2(y-b)^2+c^4}dy\;,$$ Let $(y-b) = t\;,$ Then $dy = dt$

So Integral $$\displaystyle I = \int \sqrt{4a^2t^2+c^4}dt = 2a\underbrace{\int \sqrt{t^2+k^2}dt}_{J}\;,$$ Where $\displaystyle k= \frac{c^2}{2a}$

For calculation of Integral $J\;,$ We Use Integration by parts.

Now Let $$\displaystyle J = \int \sqrt{t^2+k^2}\cdot tdt = \sqrt{t^2+k^2}\cdot t-\int\frac{t^2}{\sqrt{t^2+k^2}}dt$$

So $$\displaystyle J= \sqrt{t^2+k^2}\cdot t-\int\frac{(t^2+k^2)-k^2}{\sqrt{t^2+k^2}}dt = \sqrt{t^2+k^2}\cdot t-J+k^2\ln |t+\sqrt{t^2+k^2}|$$

So we Get $$\displaystyle J = \frac{1}{2}\sqrt{t^2+k^2}\cdot t+\frac{k^2}{2}\ln|t+\sqrt{t^2+k^2}|$$

So we get $$\displaystyle I = 2a\cdot J = a\cdot \sqrt{t^2+k^2}\cdot t+ak^2\cdot \ln|t+\sqrt{t^2+k^2}|$$

Where $\displaystyle k = \frac{c^2}{2a}$

juantheron
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HINT let $c^2\sinh\theta=2a(y-b)$

David Quinn
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Before applying Inverse Trigonometric Substitution, I prefer to preprocess the expression as follows $$\begin{array}{lll} \int\sqrt{4a^2(y-b)^2+c^4}dy &=& \displaystyle\int\sqrt{\frac{4a^2c^4(y-b)^2}{c^4}+c^4}dy\\ &=& \displaystyle \int c^2\sqrt{\frac{4a^2(y-b)^2}{c^4}+1}dy\\ &=& \displaystyle \int c^2\sqrt{\frac{(2ay-2ab)^2}{c^4}+1}dy\\ &=& \displaystyle \int c^2\sqrt{\bigg(\frac{2ay-2ab}{c^2}\bigg)^2+1}dy\\ \end{array}$$ Next, making the substitution $$u=\frac{2ay-2ab}{c^2}\implies u'(y) = \frac{2a}{c^2}$$ we have $$\int\frac{c^2\sqrt{\bigg(\frac{2ay-2ab}{c^2}\bigg)^2+1}du(y)}{\frac{2a}{c^2}} = \frac{c^4}{2a}\int \sqrt{u^2+1}du=\dots$$

John Joy
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