Question

Evaluate the integrals.

\(\int_{1/\sqrt 3 }^{\sqrt 3 } {\frac{8}{{1 + {x^2}}}} dx\)

Short answer

The value of \(\int_{1/\sqrt 3 }^{\sqrt 3 } {\frac{8}{{1 + {x^2}}}} dx\) is\(\frac{{4\pi }}{3}\).

Step-by-step solution

Definition of the integral

Integral calculus is an area of mathematics that deals with the calculation, properties, and applications of integrals.

Formula used:

\(\arctan {(x)^\prime } = \frac{1}{{1 + {x^2}}}\)

Using arctan Evaluating the integral

The given data is expressed as

\(\begin{aligned}{c}\int_{1/\sqrt 3 }^{\sqrt 3 } {\frac{8}{{1 + {x^2}}}} dx = \left. {8 \times \arctan (x)} \right|_{1/\sqrt 3 }^{\sqrt 3 }\\ = 8 \times ((\arctan (\sqrt 3 ) - \arctan (1/\sqrt 3 ))\\ = 8 \times \left( {\frac{\pi }{3} - \frac{\pi }{6}} \right)\\ = \frac{{4\pi }}{3}\end{aligned}\)

Hence the value of \(\int_{1/\sqrt 3 }^{\sqrt 3 } {\frac{8}{{1 + {x^2}}}} dx\) is\(\frac{{4\pi }}{3}\).