Question

How do you integrate $\int {\dfrac{{dx}}{{\sqrt {81 - 9{x^2}} }}}$ using trig substitutions?

Answer 1

In this method of integration by substitution, any given integral is transformed into a simple form of integral by substituting the independent variable by others. Here, to integrate the given terms using trigonometric substitution, we need to assume some trigonometric function for x such that by finding its derivative as the value of dx we can integrate the terms.

Complete step by step answer:
$\int {\dfrac{{dx}}{{\sqrt {81 - 9{x^2}} }}}$…………….. 1
As we need to integrate using trigonometric substitutions, hence
Let, $x = 3\sin u$
And its derivative, with respect to x is:
$\Rightarrow dx = 3\cos u \cdot du$
Now, substitute the obtained value of x and dx in given function i.e., equation 1 as:
$\int {\dfrac{{dx}}{{\sqrt {81 - 9{x^2}} }}}$
$\Rightarrow \int {\dfrac{{3\cos u \cdot du}}{{\sqrt {81 - 9{{\left( {3\sin u} \right)}^2}} }}}$
As the denominator consists of square terms, hence we get:
$\int {\dfrac{{3\cos u \cdot du}}{{\sqrt {81 - 9 \cdot {3^2}{{\sin }^2}u} }}}$
Simplifying the terms as: ${3^2} = 9$, we get
$\Rightarrow \int {\dfrac{{3\cos u \cdot du}}{{\sqrt {81 - 9 \cdot 9{{\sin }^2}u} }}}$
Now, multiplying the denominator terms as:
$\Rightarrow \int {\dfrac{{3\cos u \cdot du}}{{\sqrt {81 - 81{{\sin }^2}u} }}}$ ………………… 2
We know that, $\sqrt {81} = 9$, hence taking the common terms, and re-writing the equation 2 as:
$\Rightarrow \int {\dfrac{{3\cos u \cdot du}}{{9\sqrt {1 - {{\sin }^2}u} }}}$
We know that,$\sqrt {1 - {{\sin }^2}u} = \cos u$, hence we have:
$\Rightarrow \dfrac{3}{9}\int {\dfrac{{\cos u \cdot du}}{{\cos u}}}$
Simplifying the terms, we get:
$\Rightarrow \dfrac{1}{3}\int {\dfrac{{\cos u \cdot du}}{{\cos u}}}$
As, there are common terms involved i.e., $\cos u$, hence finding integration of u with respect to $du$:
$\Rightarrow \dfrac{1}{3}\int {u \cdot du}$…………….. 3
As there is a constant term involved, we know that, the integral of the constant function $\int {a \cdot du}$ is $\int {a \cdot dx = ax + c}$, hence applying this to the equation 3 with respect to $du$we get:
$\Rightarrow \dfrac{1}{3}u + c$ …………………… 4
As, we know that: $x = 3\sin u$
$\Rightarrow u = {\sin ^{ - 1}}\dfrac{x}{3}$
Hence, substitute the value of u in equation 4 as:
$\dfrac{1}{3}u + c$
$\Rightarrow \dfrac{1}{3}\left( {{{\sin }^{ - 1}}\dfrac{x}{3}} \right) + c$
or $\dfrac{1}{3}\arcsin \left( {\dfrac{x}{3}} \right) + c$

**Therefore,
$\int {\dfrac{{dx}}{{\sqrt {81 - 9{x^2}} }}} = \dfrac{1}{3}\left( {{{\sin }^{ - 1}}\dfrac{x}{3}} \right) + c$
**

Note: When a function cannot be integrated directly, then this integration by substituting method is used. 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.