Integral of 1 Over Square Root of 1-x^2

In this tutorial we shall another important integration that is integral of 1 over square root of {1-x^2}; this integral is also using as an important formula.

The integral of 1 over square root of {1-x^2} is of the form

\int  {\frac{1}{{\sqrt {1 - {x^2}} }}dx = } {\sin ^{ - 1}}x + c


To prove this formula, consider

\frac{d}{{dx}}\left[  {{{\sin }^{ - 1}}x + c} \right] = \frac{d}{{dx}}{\sin ^{ - 1}}x +  \frac{d}{{dx}}c


Using the derivative formula \frac{d}{{dx}}{\sin ^{ - 1}}x = \frac{1}{{\sqrt {1  - {x^2}} }}, we have

\begin{gathered} \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}x + c}  \right] = \frac{1}{{\sqrt {1 - {x^2}} }} + 0 \\ \Rightarrow \frac{d}{{dx}}\left[ {{{\sin }^{  - 1}}x + c} \right] = \frac{1}{{\sqrt {1 - {x^2}} }} \\ \Rightarrow \frac{1}{{\sqrt {1 - {x^2}} }} =  \frac{d}{{dx}}\left[ {{{\sin }^{ - 1}}x + c} \right] \\ \Rightarrow \frac{1}{{\sqrt {1 - {x^2}} }}dx  = d\left[ {{{\sin }^{ - 1}}x + c} \right]\,\,\,\,{\text{ -  -  -  }}\left( {\text{i}} \right) \\ \end{gathered}


Integrating both sides of equation (i) with respect to x, we have

\int  {\frac{1}{{\sqrt {1 - {x^2}} }}dx}  =  \int {d\left[ {{{\sin }^{ - 1}}x + c} \right]}


As we know that by definition integration is the inverse process of derivative, so we have

\int  {\frac{1}{{\sqrt {1 - {x^2}} }}dx}  =  {\sin ^{ - 1}}x + c


As we know that derivative of

\frac{d}{{dx}}{\cos ^{ - 1}}x =  - \frac{1}{{\sqrt {1 - {x^2}} }}

This formula can also be written as

\int  {\frac{1}{{\sqrt {1 - {x^2}} }}dx}   =  - {\cos ^{ - 1}}x + c

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