Solved Exercise: Integral Example 2

Solve the indefinite integral:

$$ \int \frac{1}{x \sqrt{1-\log^2 x}} \ dx $$

To solve the integral, use the substitution method:

$$ d( \log x ) = \frac{1}{x} \ dx $$

Multiply both sides by \( x \):

$$ d( \log x ) \cdot x = \frac{1}{x} \ dx \cdot x $$

This way, you get \( dx \):

$$ d( \log x ) \cdot x = dx $$

Substitute \( dx = d(\log x) \cdot x \) in the integral:

$$ \int \frac{1}{x \sqrt{1-\log^2 x}} \cdot [ d( \log x ) \cdot x ] $$

Simplify \( x \) from the numerator and denominator:

$$ \int \frac{1}{\sqrt{1-\log^2 x}} \cdot d( \log x ) $$

Let \( t = \log x \) as an auxiliary variable:

$$ \int \frac{1}{\sqrt{1-t^2}} \ dt $$

Now the integral is immediate:

$$ \int \frac{1}{\sqrt{1-t^2}} \ dt = \arcsin(t) + c $$

Replace \( t = \log x \):

$$ \int \frac{1}{\sqrt{1-t^2}} \ dt = \arcsin( \log x ) $$

Thus, the solution to the integral is:

$$ \int \frac{1}{x \sqrt{1-\log^2 x}} \ dx = \arcsin( \log(x) ) $$