Question

Resolva a integral ∫_0^3 (3/√9-x^2)dx
a. 3pi/2
b. 6+3pi/2
c. 3+3pi/2
d. 6+pi/2
e. 6+pi


r

Answer 1

$\int_{0}^3 \frac{3}{\sqrt{9-x^2}} dx = \boxed{\boxed{3\int_{0}^3 \frac{1}{\sqrt{9-x^2}} dx}}$

a substituição trigonometrica  indicada para esse caso

$\boxed{\boxed{\sqrt{a^2-x^2} \to x=asen(\theta)}}}$

aplicando isso
$\boxed{x=3sen(\theta)} \\\\ \frac{dx}{d\theta} = 3cos(\theta)\\\\ \boxed{ dx= 3cos(\theta) d\theta}$


substituindo na integral

$3\int_{0}^3 \frac{1}{\sqrt{9-(3sen(\theta))^2}} *3cos(\theta)\; d\theta \\\\ = 3*3 \int_{0}^3 \frac{cos(\theta)}{\sqrt{9-9sen^2(\theta))}} d\theta\\\\ = 9\int_0^3 \frac{cos(\theta)}{\sqrt{9*(1-sen^2(\theta)}} \;d\theta\\\\ = \frac{9}{\sqrt{9}} \int_0^3 \frac{cos(\theta)}{\sqrt{cos^2(\theta)}} \;d\theta\\\\ = 3 \int_0^3 \frac{cos(\theta)}{cos(\theta)} d\theta\\\\ =3\int _0^3 d\theta = 3 \left[\theta \right]^3_0 = 3\left[arcsen( \frac{x}{3}) \right]^3_0 = 3 [arcsen( \frac{3}{3})-arcsen(0)]$

$\\\\ = \frac{3\pi}{2}$