Question

How do you integrate $\int \dfrac{x}{{\sqrt {{x^2} - 7} }}dx$ by trigonometric substitution?

Answer 1

To solve this question, first we will assume any of the set of variables or constant be another variable to get the expression easier. And conclude until the non-operational state is not achieved. And finally substitute the assumed value. It is also called integration by substitution.

Complete step by step solution:
The given expression: $\int \dfrac{x}{{\sqrt {{x^2} - 7} }}dx$
We can integrate this expression by the substitution-
Let ${x^2} - 7 = t$ .
Now, differentiate the above assumed equation:
$\Rightarrow \dfrac{{d{x^2}}}{{dx}} - \dfrac{d}{{dx}}(7) = \dfrac{{dt}}{{dx}}$
$\Rightarrow 2x - 0 = \dfrac{{dt}}{{dx}}$
$\Rightarrow 2x.dx = dt$
$\Rightarrow x.dx = \dfrac{{dt}}{2}$
Now, use the above equation in the main expression:
$\because \int \dfrac{x}{{\sqrt {{x^2} - 7} }}dx \;$
put $\dfrac{{dt}}{2}$ instead of $x.dx$ .
$= \int \dfrac{{dt}}{2}\dfrac{1}{{\sqrt t }} \\\ = \dfrac{1}{2}\int {t^{ - \dfrac{1}{2}}}dt \\\ = \dfrac{1}{2}\dfrac{{{t^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}} + C \\\ = \sqrt t + C \;$
Now, substitute the actual value of $t$ :
$= \sqrt {{x^2} - 7} + C$
Hence, the integration of
$\int \dfrac{x}{{\sqrt {{x^2} - 7} }}dx \;$ is $\sqrt {{x^2} - 7} + C$ .
So, the correct answer is “ $\sqrt {{x^2} - 7} + C$ ”.

Note : Usually the method of integration by substitution is extremely useful when we make a substitution for a function whose derivative is also present in the integrand. Doing so, the function simplifies and then the basic formulas of integration can be used to integrate the function.