Evaluate the integral.
$$\int \frac { 6 \sin 2 x } { 1 + \cos ^ { 4 } x } d x$$
A) $6 \arctan \left( \sec ^ { 2 } x \right) + C$
B) $- 6 \cos ^ { 2 } 2 x + C$
C) $- 6 \arcsin ( \tan x ) + C$
D) $1 - 6 \cos ^ { 2 } x + C$
E) None of these

Question

Quizplus · Accepted Answer

The integral can be evaluated using the substitution $u = \cos^2 x$. Then, $du = -2\sin 2x dx$, and the integral becomes:$$\int \frac { 6 \sin 2 x } { 1 + \cos ^ { 4 } x } d x = -3 \int \frac { 1 } { 1 + u ^ { 2 } } d u = -3 \arctan u + C = -3 \arctan (\cos^2 x) + C$$

Evaluate the Integral ∫6sin⁡2x1+cos⁡4xdx\int \frac { 6 \sin 2 x } { 1 + \cos ^ { 4 } x } d x∫1+cos4x6sin2x​dx

Evaluate the Integral $\int \frac { 6 \sin 2 x } { 1 + \cos ^ { 4 } x } d x$