Evaluate the Integral Using the Indicated Trigonometric Substitution $\int \frac{x^{3}}{\sqrt{x^{2}+16}} d x ; x=4 \tan \theta$

Evaluate the integral using the indicated trigonometric substitution. $\int \frac{x^{3}}{\sqrt{x^{2}+16}} d x ; x=4 \tan \theta$

A) $\frac{3}{2}(x+16) ^{3 / 2}-16 \sqrt{x+16}+C$
B) $\left(x^{2}+16\right) ^{3 / 2}-\sqrt{x^{2}+16}+C$
C) $\frac{1}{3}\left(x^{2}+16\right) ^{3 / 2}-\sqrt{x^{2}+16}+C$
D) $\frac{1}{3}\left(x^{2}+16\right) ^{3 / 2}+16 \sqrt{x^{2}+16}+C$
E) $\left(x^{2}+16\right) ^{3 / 2}-4 \sqrt{x^{2}+16}+C$

Correct Answer:

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Access For Free