Question

How do you integrate \[\dfrac{1}{{\sqrt {1 - {x^2}} }}\]?

Answer 1

Hint: In the given question, we have been given an algebraic expression which is to be integrated. The expression consists of a square root bracket with two terms in it – a constant and a variable raised to the second power. The two terms are then separated by a minus sign. We have to integrate this whole expression. To do that, we are going to replace the square root bracket by expressing it in the form of a power and then applying the integration formula to solve it.

Complete step-by-step answer:
The given expression is \[\dfrac{1}{{\sqrt {1 - {x^2}} }}\].
We are first going to solve the expression without the fraction, i.e., we are going to integrate \[\sqrt {1 - {x^2}} \] and then take the reciprocal of the result.
Let \[x = \sin \theta \]
Then \[dx = \cos \theta d\theta \]
Now, \[\cos \theta = \sqrt {1 - {x^2}} \]
We know, \[\sin 2\theta = 2\sin \theta \cos \theta \]
Hence, \[\sin 2\theta = 2x\sqrt {1 - {x^2}} \]
Thus, the integral is
\[I = \int {\sqrt {1 - {x^2}} dx = \int {\cos \theta .\cos \theta d\theta } } = \int {{{\cos }^2}\theta d\theta } \]
Now, we know
\[\cos 2\theta = 2{\cos ^2}\theta - 1 \Rightarrow {\cos ^2}\theta = \dfrac{{1 + \cos 2\theta }}{2}\]
Therefore,
\[I = \dfrac{1}{2}\int {\left( {1 + \cos 2\theta } \right)d\theta } \]
\[ = \dfrac{1}{2}\left( {\theta + \sin 2\theta } \right) + C\]
We know that, \[\theta = \arcsin x\]
Hence,
\[I = \dfrac{1}{2}\arcsin x + \dfrac{1}{2}x\sqrt {1 - {x^2}} + C\]
Now, we take the reciprocal,
\[I' = \dfrac{1}{I} = \dfrac{1}{{\dfrac{1}{2}\arcsin x + \dfrac{1}{2}x\sqrt {1 - {x^2}} + C}} = \dfrac{2}{{\arcsin x + x\sqrt {1 - {x^2}} }} + C\]

Additional Information:
Integration is the opposite of differentiation. In differentiation, we “break” things for examining how they behave separately. While, in integration, we combine the expressions so as to see their collective behavior. If we have a definite integral, then we calculate its value by putting in the upper limit into the result, then putting in the lower limit into the result and then subtracting the two.


Note: In the given question, we had been given an algebraic expression which was to be integrated. To do that, we replaced the square root bracket by expressing it in the form of a power and then applied the integration formula to solve it. It is thus, really important that we know the formula of the integration, how to transform one quantity (algebra) to another (trigonometric) and how to use it in this case. Care must be taken when we are performing the steps as that is the point where the things get tricky and there are chances of getting it wrong.