Answered: "Evaluate the integrals ∫33dt3+t2"

Dillard

Answered question

2021-10-20

Evaluate the integrals

\int_{\sqrt{3}}^{3} \frac{d t}{3 + t^{2}}

Alannej

Skilled2021-10-21Added 104 answers

We have to evaluate the integral

\int_{\sqrt{3}}^{3} \frac{d t}{3 + t^{2}}

We know

\int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a})

Therefore,

\int \frac{d x}{x^{2} + a^{2}} = \frac{1}{a} \tan^{- 1} (\frac{x}{a})

Therefore,

\int_{\sqrt{3}}^{3} = \int_{\sqrt{3}}^{3} \frac{d t}{t^{2} + {(\sqrt{3})}^{2}}

= [\frac{1}{\sqrt{3}} \tan^{- 1} {(\frac{t}{\sqrt{3}}]}_{\sqrt{3}}^{3}

= [\frac{1}{\sqrt{3}} \frac{π}{3} - \frac{1}{\sqrt{3}} \tan^{- 1} (1)]

= [\frac{1}{\sqrt{3}} (\frac{π}{3} - \frac{π}{4}]

= \frac{1}{\sqrt{3}} \frac{π}{12}

= \frac{π}{12 \sqrt{3}}

= \frac{\sqrt{3} π}{36}

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