Expert answers to "Evaluate the integrals. ∫22dxxx2−1" - Plainmath

Nann

Answered question

2021-10-10

Evaluate the integrals.

\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2} - 1}}

Answer & Explanation

Brighton

Skilled2021-10-11Added 103 answers

Evaluate,

\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2} - 1}}

Simplification:
Since we have,

\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2} - 1}}

Substitute

\sqrt{x^{2} - 1} = u \Rightarrow \frac{2 x}{2 \sqrt{x^{2} - 1}} d x = d u \Rightarrow \frac{1}{\sqrt{x^{2} - 1}} d x = \frac{1}{x} d u

When

x = \sqrt{2}, u = 1

When

x = 2, u = 3

In (1),

\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2} - 1}} = \int_{1}^{\sqrt{3}} \frac{d u}{x^{2}}

= \int_{1}^{\sqrt{3}} \frac{d u}{x^{2}}

= \int_{1}^{\sqrt{3}} \frac{d u}{1 + u^{2}}

= {[\tan^{- 1} u]}_{1}^{\sqrt{3}}

= [\tan^{- 1} \sqrt{3} - \tan^{- 1} 1]

= [\frac{π}{3} - \frac{π}{4}]

= [\frac{4 π - 3 π}{12}]

= \frac{π}{12}

Hence

\int_{\sqrt{2}}^{2} \frac{d x}{x \sqrt{x^{2} - 1}} = \frac{π}{12}

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