Question

23–68. Indefinite integrals Determine the following indefinite integrals. Check your work by differentiation.∫ (1/x√(36x² - 36))dx

Accepted Answer

Rewrite the integral in a more manageable form. Factor out constants from the square root in the denominator: $$ \int \frac{1}{x \sqrt{36x^2 - 36}} \, dx . $$Factor out $$36$$ from the square root: $$ \sqrt{36x^2 - 36} = \sqrt{36(x^2 - 1)} = 6\sqrt{x^2 - 1} . $$The integral becomes $$ \int \frac{1}{6x\sqrt{x^2 - 1}} \, dx . $$Recognize the structure of the integrand. The term $$ \sqrt{x^2 - 1} $$ suggests a trigonometric substitution. Use the substitution $$ x = \sec(	heta) $$, which implies $$ dx = \sec(	heta)	an(	heta) \, d	heta $$ and $$ \sqrt{x^2 - 1} = 	an(	heta) . $$Substitute $$ x = \sec(	heta) $$ into the integral. The integral becomes $$ \int \frac{1}{6\sec(	heta)	an(	heta) \cdot \sec(	heta)} \sec(	heta)	an(	heta) \, d	heta . $$Simplify the expression: $$ \int \frac{1}{6\sec^2(	heta)} \, d	heta = \frac{1}{6} \int \cos^2(	heta) \, d	heta . $$Simplify further using the trigonometric identity $$ \cos^2(	heta) = \frac{1 + \cos(2	heta)}{2} . $$The integral becomes $$ \frac{1}{6} \int \frac{1 + \cos(2	heta)}{2} \, d	heta = \frac{1}{12} \int (1 + \cos(2	heta)) \, d	heta . $$Split the integral: $$ \frac{1}{12} \int 1 \, d	heta + \frac{1}{12} \int \cos(2	heta) \, d	heta . $$Evaluate each term. The first term is $$ \frac{1}{12} 	heta . $$For the second term, use the integral of $$ \cos(2	heta) $$, which is $$ \frac{1}{2}\sin(2	heta) . $$Combine the results: $$ \frac{1}{12} 	heta + \frac{1}{24}\sin(2	heta) + C . $$Finally, back-substitute $$ 	heta $$ using $$ x = \sec(	heta) $$, where $$ 	heta = \sec^{-1}(x) $$, to express the solution in terms of $$ x .$$

23–68. Indefinite integrals Determine the following indefinite integrals. Check your work by differentiation.

∫ (1/x√(36x² - 36))dx

Verified video answer for a similar problem:

Key Concepts

Indefinite Integrals

Substitution Method

Differentiation Check