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I need help with the proof of the table of integral that: $$\int \frac{\sqrt{a^2 + u^2}}{u} du = \sqrt{a^2 + u^2} + a \ln \left|\frac{\sqrt{a^2 + u^2} - a}{u}\right| + c$$

Must solve using trigonometric substitution.

I understand that you have to use $a^2\tan(\theta)$ as a substitution for $u$.

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  • $\begingroup$ Is this what you meant? If not, you can use MathJax to change it yourself. $\endgroup$
    – Toby Mak
    Feb 4, 2019 at 13:32
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    $\begingroup$ take the derivation of the right side and compare it with the left? $\endgroup$
    – Bonbon
    Feb 4, 2019 at 13:48
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    $\begingroup$ Why must you use trig substitution? Of course, to prove something in the integral table, you merely have to differentiate the right hand side and verify that it is the integrand. $\endgroup$
    – GEdgar
    Feb 5, 2019 at 1:30

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Let $u=a \tan x \tag{1}$ Then:- $u^2=a^2\tan^2 x \implies 2u \cdot du= 2a^2\tan x \cdot \sec^2x \cdot dx $

The integral reduces to :- $$ a\int \sec^2x \csc x \cdot dx $$ $$ = a\int (\tan^2x +1) \csc x \cdot dx = a\left(\int \sec x \tan x \cdot dx + \int \csc x \cdot dx \right)\\ = a\left(\sec x - \ln(\cot x + \csc x)\right) + C $$Note that from $(1)$ we have $\sec x = \frac{\sqrt{u^2+a^2}}{a}$ and $\csc x= \frac{\sqrt{u^2+a^2}}{u}$ and $\cot x = \frac{a}{u} $

Thus , by undoing substitution , we get $$\boxed{ I= \sqrt{a^2+u^2} + a\cdot \ln \left|\frac{\sqrt{a^2+u^2}-a}{u}\right| + C}$$

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  • $\begingroup$ Thanks for your help! $\endgroup$ Feb 5, 2019 at 16:37

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