Explained: "Trigonometric substitutions Evaluate the following integrals using trigonometric..."

FobelloE

Answered question

2021-10-07

Trigonometric substitutions Evaluate the following integrals using trigonometric substitution.

\int \frac{d x}{x^{3} \sqrt{x^{2} - 1}}, x > 1

Alannej

Skilled2021-10-08Added 104 answers

Consider the provided integral,

\int \frac{d x}{x^{3} \sqrt{x^{2} - 1}}, x > 1

Evaluate the following integrals using trigonometric substitution.
Apply the trigonometric substitution let

x = \sec (u) \Rightarrow d x = \sec (u) \tan (u) d u

So,

\int \frac{d x}{x^{3} (\sqrt{x^{2} - 1})} = \int \frac{\sec (u) \tan (u) d u}{\sec^{3} (u) \sqrt{\sec^{2} (u) - 1}}

= \int \frac{\tan (u)}{\sec^{2} (u) \tan (u)} d u

= \int \frac{1}{\sec^{2} (u)} d u

Using trigonometric identity simplifying further,

\int \frac{d x}{x^{3} \sqrt{x^{2} - 1}} = \int \cos^{2} (u) d u

= \int \frac{1 + \cos (u)}{2} d u

= \frac{1}{2} (\int 1 d u + \int \cos (2 u) d u)

= \frac{1}{2} (u + \frac{1}{2} \sin (2 u)) + C

Substitute back

x = \sec (u)

\int \frac{d x}{x^{3} \sqrt{x^{2} - 1}} = \frac{1}{2} (a r c \sec (x) + \frac{1}{2} a r c \sec (x)) + C

Hence.

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