Question

Evaluate the definite integral. $$ \int_{2}^{4} \sqrt{3+x^{2}} d x $$

Short answer

The result of the integral is \( 2\sqrt{3} [\frac{1}{2} (\sec(4)\tan(4) + \ln | \sec(4) + \tan(4) |) - \frac{1}{2} (\sec(2)\tan(2) + \ln | \sec(2) + \tan(2) |)] \)

Step-by-step solution

Trigonometric Substitution

Assign \( x = \sqrt{3} \tan(\theta) \) which implies \( dx = \sqrt{3} \sec^2(\theta) d \theta \). This substitution is suggested because the square root in the integrand will simplify.

Substitute and Simplify

Substitute \( x = \sqrt{3} \tan(\theta) \) and \( dx = \sqrt{3} \sec^2(\theta) d \theta \) into the integral: \( \int_{2}^{4} \sqrt{3 + x^2} d x = \int_{\arctan(2 / \sqrt{3})}^{\arctan(4 / \sqrt{3})}\sqrt{3 + 3 \tan^2(\theta)} \sqrt{3} \sec^2(\theta) d \theta \). Then, simplify the integrand, which becomes \( 2\sqrt{3} \int_{\arctan(2 / \sqrt{3})}^{\arctan(4 / \sqrt{3})} \sec^3(\theta) d \theta \)

Evaluate the Integral

This integral is now a standard one that is evaluated as follows: \( 2\sqrt{3} [\frac{1}{2} (\sec(\theta)\tan(\theta) + \ln | \sec(\theta) + \tan(\theta) |)]_{\arctan(2 / \sqrt{3})}^{\arctan(4 / \sqrt{3})} \)

Substitute the Limits

Substitute back the values using the arctan limits to the above result: \( 2\sqrt{3} [\frac{1}{2} (\sec(\theta)\tan(\theta) + \ln | \sec(\theta) + \tan(\theta) |)]_{2}^{4} \)

Simplify the final result

By simplifying, the final answer would be \( 2\sqrt{3} [\frac{1}{2} (\sec(4)\tan(4) + \ln | \sec(4) + \tan(4) |) - \frac{1}{2} (\sec(2)\tan(2) + \ln | \sec(2) + \tan(2) |)] \)