Evaluate the integral by using multiple substitutions.
-$$\int \frac { 6 \tan ^ { 2 } x \sec ^ { 2 } x } { \left( 3 + \tan ^ { 3 } x \right) ^ { 2 } } d x$$
A) $\frac { 2 \tan ^ { 3 } x \sec ^ { 3 } x } { 9 + \tan ^ { 3 } x } + C$
B) $\frac { - 2 } { 3 + \tan ^ { 3 } x } + C$
C) $\frac { 6 x ^ { 2 } } { \left( 3 + x ^ { 3 } \right) ^ { 2 } } + C$
D) $\frac { 2 } { 3 + \cot ^ { 3 } x } + C$

Question

Quizplus · Accepted Answer

To solve the integral, let $u = 3 + \tan^3x$, then $du = 3\tan^2x\sec^2x dx$. Adjusting the integral to match this substitution, we get $\int \frac{2}{u^2} du$, which simplifies to $\frac{-2}{u} + C$ or $\frac{-2}{3 + \tan^3x} + C$.

Evaluate the Integral by Using Multiple Substitutions ∫6tan⁡2xsec⁡2x(3+tan⁡3x)2dx\int \frac { 6 \tan ^ { 2 } x \sec ^ { 2 } x } { \left( 3 + \tan ^ { 3 } x \right) ^ { 2 } } d x∫(3+tan3x)26tan2xsec2x​dx

Evaluate the Integral by Using Multiple Substitutions $\int \frac { 6 \tan ^ { 2 } x \sec ^ { 2 } x } { \left( 3 + \tan ^ { 3 } x \right) ^ { 2 } } d x$