# Evaluate the integral {eq}\int \frac{1}{\sqrt{x^2-16}}\mathrm{d}x {/eq} using trigonometric substitution.

## Question:

Evaluate the integral {eq}\int \frac{1}{\sqrt{x^2-16}}\mathrm{d}x {/eq} using trigonometric substitution.

## Trigonometric Substitution:

The trigonometric substitutions that are effective for the given radical expressions because of the specified trigonometric identities are given as:

{eq}\displaystyle{ \left. \begin{array} { l l } { \sqrt { a ^ { 2 } - x ^ { 2 } } } & { x = a \operatorname { sin } \theta } \\ { \sqrt { a ^ { 2 } + x ^ { 2 } } } & { x = a \operatorname { tan } \theta } \\ { \sqrt { x ^ { 2 } - a ^ { 2 } } } & { x = a \operatorname { sec } \theta } \end{array} \right. }{/eq}

Important identity:

{eq}\displaystyle{ \operatorname { sec } ^ { 2 } \theta - 1 = \operatorname { tan } ^ { 2 } \theta }{/eq}

and the integral of:

{eq}\displaystyle{ \int \operatorname { sec } x d x = \operatorname { ln } | \operatorname { sec } x + \operatorname { tan } x | + C }{/eq}

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We have,

{eq}\displaystyle{ \int \frac{1}{\sqrt{x^2-16}}\mathrm{d}x }{/eq}

Substitute {eq}\displaystyle{ x = 4 \operatorname { sec } \theta ...