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# What Euler’s substitutions must be used in integrals of the form: $\sqrt {a{x^2} + bx + c}$ $\pm \sqrt b x,a > 0$  $\pm \sqrt {{a^3}} x,a > 0$  $\pm \sqrt a x,a > 0$ None of These

Last updated date: 20th Jun 2024
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Hint: In this question, we need to determine the Euler’s substitution that must be used in evaluating the integrals of the type $\sqrt {a{x^2} + bx + c}$. Use all the options in Euler’s substitutions methods to get the desired results and check which of the options is best suitable.
Formula Used: As such there is no formula used. Just to get rid of the square root and quadratic equation, we use Euler’s substitutions methods.

For option A: Our assumption is $a > 0$ .
Substitute: $\sqrt {a{x^2} + bx + c} = x\sqrt b + t$
Squaring both sides,
We get, $a{x^2} + bx + c = {x^2}b + {t^2} + 2x\sqrt b$
$a{x^2} - {x^2}b + bx - 2x\sqrt b = {t^2} - c$
On taking out x common from left hand side,
We get, $x\left( {ax - xb + b - 2\sqrt b } \right) = {t^2} - c$
Hence we cannot solve further as we have to solve x in terms of constants which is not possible.

For option B: Our assumption is $a > 0$ .
Substitute: $\sqrt {a{x^2} + bx + c} = x\sqrt {{a^3}} + t$
Squaring both sides,
We get, $a{x^2} + bx + c = {x^2}{a^3} + {t^2} + 2x\sqrt {{a^3}}$
$a{x^2} - {x^2}{a^3} + bx - 2x\sqrt {{a^3}} = {t^2} - c$
On taking out x common from left hand side,
We get, $x\left( {ax - x{a^3} + b - 2\sqrt {{a^3}} } \right) = {t^2} - c$
Hence we cannot solve further as we have to solve x in terms of constants which is not possible.

For option C: Our assumption is $a > 0$ .
Substitute: $\sqrt {a{x^2} + bx + c} = x\sqrt a + t$
Squaring both sides,
We get, $a{x^2} + bx + c = {x^2}a + {t^2} + 2x\sqrt a$
Cancelling $a{x^2}$ on both sides,
We get, $bx + c = {t^2} + 2x\sqrt a$
$bx - 2x\sqrt a = {t^2} - c$
On taking out x common from left hand side,
We get, $x\left( {b - 2\sqrt a } \right) = {t^2} - c$
So, $x$ comes out to be: $x = \dfrac{{{t^2} - c}}{{b - 2\sqrt a }}$
Hence, we calculate x in terms of constants for $a > 0$ .
So, option C is correct, that is $\pm \sqrt a x,a > 0$ .
So, the correct answer is “Option C”.

Note: Here we use Euler’s substitutions to check whether x comes out to be in constants for $a > 0$ .As to get rid of the square roots and quadratic equation. So, we verified all the options to get the desired results. We have three Euler’s substitutions for different values of $\left( {a,b,c} \right)$ . First we assume $a > 0$ , for second we assume $b \ne 0$ and third we assume $c > 0$ .