can someone share the proof for this integrals (1 Viewer)

ExtremelyBoredUser · Jun 11, 2022

SCREW L+S said:
View attachment 35777

You can just consider the derivatives of the answer and reverse engineer (Reverse chain rule). which is way more easier (just manipulate the fraction), but I'll treat it like I have no foresight:

$\int \frac{1}{\sqrt{x^2+a^2}} \, dx$
Let $x = a\tan(u)$ so $dx = a\sec^2(u)$

$\int \frac{1}{\sqrt{a^{2}\times(\tan^2(u) + 1)}} \times a\sec^2(u) \, du$
$\int \frac{\sec^2(u)}{\sqrt{a^{2}\times(sec^2(u))}} \, du$
$\int \frac{a\sec^2(u)}{a\sec(u)} \, du$
$\int \frac{a\sec(u)}{a} \, du$
$\int \sec(u) \, du$
$\ln{|\tan(u) + \sec(u)|}$
$\ln{|\frac{x}{a} + \frac{\sqrt{a^2+x^2}}{a}|}$
$\ln{|\frac{x+\sqrt{a^2+x^2}}{a}|}$
$\ln{|x+\sqrt{a^2+x^2}|} - ln(a)$
$\ln{|x+\sqrt{a^2+x^2|}+ C$ as ln(a) is a constant value

Similarly for 2, just let x = asecu so you get something similar

stupid_girl · Jun 12, 2022

What's the point to create two separate formula when you can simply replace a^2 and -a^2 by a real constant k to get a single formula?

Differentiating RHS will give you LHS readily.

can someone share the proof for this integrals (1 Viewer)

SCREW L+S

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ExtremelyBoredUser

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stupid_girl

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