evaluate the following integral using trigonometric substitution.\n int_{0}^{2sqrt{2}} \frac{x^{2}}{sqrt{16…

evaluate the following integral using trigonometric substitution.\n int_{0}^{2sqrt{2}} \frac{x^{2}}{sqrt{16 - x^{2}}} dx \nwhat substitution will be the most helpful for evaluating this integral?\na. (x = 4\tan\theta)\nb. (x = 4sec\theta)\nc. (x = 4sin\theta)

evaluate the following integral using trigonometric substitution.\n int_{0}^{2sqrt{2}} \frac{x^{2}}{sqrt{16 - x^{2}}} dx \nwhat substitution will be the most helpful for evaluating this integral?\na. (x = 4\tan\theta)\nb. (x = 4sec\theta)\nc. (x = 4sin\theta)

Answer

Explanation:

Step1: Analyze the form of integrand

We have $\sqrt{16 - x^{2}}$ in the denominator. For expressions of the form $\sqrt{a^{2}-x^{2}}$, the substitution $x = a\sin\theta$ is useful. Here $a = 4$, so $x=4\sin\theta$.

Step2: Find the correct substitution option

Based on the above - mentioned rule, the substitution $x = 4\sin\theta$ will simplify the integral.

Answer:

C. $x = 4\sin\theta$