Question

How can you prove \[\int {\dfrac{{dx}}{{{a^2} - {x^2}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + c} \] using \[x = a\sin \theta \]?

Answer 1

Hint: In solving the given question, first derive the given condition i.e., \[x = a\sin \theta \], and we will get and trigonometric expression by using some trigonometric identities and formulas, and after simplifying the expression we will get an expression that is easily integrated and after integrating we will get the expression on the right hand side, which is the given question.

Complete step-by-step solution:
Given expression is \[\int {\dfrac{{dx}}{{{a^2} - {x^2}}}} \], we have to prove that this equal to \[\dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + c\] using \[x = a\sin \theta \]
Now given that,
\[x = a\sin \theta \],
 Now differentiating both sides we get,
\[ \Rightarrow dx = \dfrac{d}{{dx}}a\sin \theta \],
Applying derivatives we get,
\[ \Rightarrow dx = a\cos \theta d\theta \],
Now substituting the values we get in the given expression \[\int {\dfrac{{dx}}{{{a^2} - {x^2}}}} \], we get,
\[ \Rightarrow \int {\dfrac{{a\cos \theta d\theta }}{{{a^2} - {{\left( {a\sin \theta } \right)}^2}}}} \],
Now simplifying we get,
\[ \Rightarrow \int {\dfrac{{a\cos \theta d\theta }}{{{a^2} - {a^2}{{\sin }^2}\theta }}} \],
Now taking out the common terms, we get,
\[ \Rightarrow \int {\dfrac{{a\cos \theta d\theta }}{{{a^2}\left( {1 - {{\sin }^2}\theta } \right)}}} \],
Now using the trigonometric identities i.e.,\[1 - {\sin ^2}\theta = {\cos ^2}\theta \], then the expression becomes,
\[ \Rightarrow \int {\dfrac{{a\cos \theta d\theta }}{{{a^2}{{\cos }^2}\theta }}} \],
Now eliminating the like terms we get,
\[ \Rightarrow \int {\dfrac{{d\theta }}{{a\cos \theta }}} \],
Now again using trigonometric identity\[\cos \theta = \dfrac{1}{{\sec \theta }}\]we get,
\[ \Rightarrow \dfrac{1}{a}\int {\sec \theta d\theta } \],
Now multiplying and dividing with\[\sec \theta + \tan \theta \], we get,
\[ \Rightarrow \dfrac{1}{a}\int {\dfrac{{\sec \theta \left( {\sec \theta + \tan \theta } \right)}}{{\sec \theta + \tan \theta }}d\theta } \],
Now simplifying we get,
\[ \Rightarrow \dfrac{1}{a}\int {\dfrac{{{{\sec }^2}\theta + \sec \theta \tan \theta }}{{\sec \theta + \tan \theta }}d\theta } \],
Now put u=\[\sec \theta + \tan \theta \],
Now derivating on both sides we get,
\[du = d\left( {\sec \theta + \tan \theta } \right)\],
Derivating each term we get,
\[du = {\sec ^2}\theta + \sec \theta \tan \theta d\theta \],
Now substituting we get,
\[ \Rightarrow \dfrac{1}{a}\int {\dfrac{{du}}{u}} \],
Now applying integration we get,
\[ \Rightarrow \dfrac{1}{a}\left| {\log u} \right|\],
Now we know that u=\[\sec \theta + \tan \theta \],,
So, substituting the value we get,
\[ \Rightarrow \dfrac{1}{a}\left| {\log \left( {\sec \theta + \tan \theta } \right)} \right|\],
Now using trigonometric identities \[\sec \theta = \dfrac{1}{{\cos \theta }},\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\], we get,
\[ \Rightarrow \dfrac{1}{a}\left| {\log \left( {\dfrac{1}{{\cos \theta }} + \dfrac{{\sin \theta }}{{\cos \theta }}} \right)} \right|\],
Now simplifying we get,
\[ \Rightarrow \dfrac{1}{a}\left| {\log \left( {\dfrac{{1 + \sin \theta }}{{\cos \theta }}} \right)} \right|\],
Now multiplying and dividing with 2, we get,
\[ \Rightarrow \dfrac{1}{{2a}}\left| {2\log \left( {\dfrac{{1 + \sin \theta }}{{\cos \theta }}} \right)} \right|\],
Now using logarithmic properties we get,
\[ \Rightarrow \dfrac{1}{{2a}}\left| {\log {{\left( {\dfrac{{1 + \sin \theta }}{{\cos \theta }}} \right)}^2}} \right|\]
Now by simplifying we get,
\[ \Rightarrow \dfrac{1}{{2a}}\left| {\log \left( {\dfrac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{{{\cos }^2}\theta }}} \right)} \right|\],
Now again using trigonometric identities, \[{\cos ^2}\theta = 1 - {\sin ^2}\theta \] we get,
\[ \Rightarrow \dfrac{1}{{2a}}\left| {\log \left( {\dfrac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{1 - {{\sin }^2}\theta }}} \right)} \right|\],
Now simplifying we get,
\[ \Rightarrow \dfrac{1}{{2a}}\left| {\log \left( {\dfrac{{{{\left( {1 + \sin \theta } \right)}^2}}}{{\left( {1 - \sin \theta } \right)\left( {1 + \sin \theta } \right)}}} \right)} \right|\],
Now eliminating the like terms we get,
\[ \Rightarrow \dfrac{1}{{2a}}\left| {\log \left( {\dfrac{{\left( {1 + \sin \theta } \right)}}{{\left( {1 - \sin \theta } \right)}}} \right)} \right|\]
Now using the condition \[x = a\sin \theta \],
\[ \Rightarrow \sin \theta = \dfrac{x}{a}\],
Now substituting the value in the expression we get,
\[ \Rightarrow \dfrac{1}{{2a}}\left| {\log \left( {\dfrac{{\left( {1 + \dfrac{x}{a}} \right)}}{{\left( {1 - \dfrac{x}{a}} \right)}}} \right)} \right|\],
Now simplifying we get,
\[ \Rightarrow \dfrac{1}{{2a}}\left| {\log \left( {\dfrac{{\left( {\dfrac{{a + x}}{a}} \right)}}{{\left( {\dfrac{{a - x}}{a}} \right)}}} \right)} \right|\],
Now eliminating the denominators we get,
\[ \Rightarrow \dfrac{1}{{2a}}\left| {\log \left( {\dfrac{{a + x}}{{a - x}}} \right)} \right| + c\]
Hence proved.

When we use the condition \[x = a\sin \theta \], then \[\int {\dfrac{{dx}}{{{a^2} - {x^2}}} = \dfrac{1}{{2a}}\log \left| {\dfrac{{a + x}}{{a - x}}} \right| + c} \].
Hence proved.

Note: In some of the integration questions, trigonometric identities are used to simplify any integral function which consists of trigonometric functions. It simplifies the integral function so that it can be easily integrated. There are many trigonometric identities, here are some useful identities:
 \[{\sin ^2}x = 1 - {\cos ^2}x\],
\[{\cos ^2}x = 1 - {\sin ^2}x\],
\[{\sec ^2}x - {\tan ^2}x = 1\],
\[{\csc ^2}x = 1 + {\cot ^2}x\].