Hitunglah hasil dari integral 2 4 akar(x^2-4)/x dx.

2\sqrt{3} - \frac{2\pi}{3}

Pembahasan

Untuk menghitung integral dari \int_{2}^{4} \frac{\sqrt{x^2-4}}{x} dx, kita dapat menggunakan substitusi trigonometri. Misalkan x = 2 sec \theta, maka dx = 2 sec \theta tan \theta d\theta.

Ketika x = 2, 2 = 2 sec \theta => sec \theta = 1 => \theta = 0.
Ketika x = 4, 4 = 2 sec \theta => sec \theta = 2 => \theta = \frac{\pi}{3}.

Substitusi ke dalam integral:
\int_{0}^{\frac{\pi}{3}} \frac{\sqrt{(2 \sec \theta)^2-4}}{2 \sec \theta} (2 \sec \theta \tan \theta) d\theta
= \int_{0}^{\frac{\pi}{3}} \frac{\sqrt{4 \sec^2 \theta-4}}{2 \sec \theta} (2 \sec \theta \tan \theta) d\theta
= \int_{0}^{\frac{\pi}{3}} \frac{\sqrt{4(\sec^2 \theta-1)}}{2 \sec \theta} (2 \sec \theta \tan \theta) d\theta
= \int_{0}^{\frac{\pi}{3}} \frac{\sqrt{4 \tan^2 \theta}}{2 \sec \theta} (2 \sec \theta \tan \theta) d\theta
= \int_{0}^{\frac{\pi}{3}} \frac{2 \tan \theta}{2 \sec \theta} (2 \sec \theta \tan \theta) d\theta
= \int_{0}^{\frac{\pi}{3}} 2 \tan^2 \theta d\theta
= 2 \int_{0}^{\frac{\pi}{3}} (\sec^2 \theta - 1) d\theta
= 2 [\tan \theta - \theta]_{0}^{\frac{\pi}{3}}
= 2 [(\tan \frac{\pi}{3} - \frac{\pi}{3}) - (\tan 0 - 0)]
= 2 [(\sqrt{3} - \frac{\pi}{3}) - (0)]
= 2\sqrt{3} - \frac{2\pi}{3}