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

Hint: 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.