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

Using the trigonometric identity $1 + \cos ^ { 4 } x = 2 \cos ^ { 2 } x + \cos ^ { 4 } x = ( \cos ^ { 2 } x + 1 ) ( \cos ^ { 2 } x ) = \sec ^ { 2 } x \tan ^ { 2 } x$, we can rewrite the integral as$$\int \frac { 6 \sin 2 x } { 1 + \cos ^ { 4 } x } d x = \int \frac { 6 \sin 2 x } { \sec ^ { 2 } x \tan ^ { 2 } x } d x = \int \frac { 6 \sin 2 x \cos ^ { 2 } x } { \tan ^ { 2 } x } d x$$Now, let $u = \tan x$. Then $du = \sec ^ { 2 } x dx$, and the integral becomes$$\int \frac { 6 \sin 2 x \cos ^ { 2 } x } { \tan ^ { 2 } x } d x = \int \frac { 6 \sin 2 x } { u ^ { 2 } } d u = - 6 \int \frac { \sin 2 x } { u ^ { 2 } } d u$$Using the trigonometric identity $\sin 2 x = 2 \sin x \cos x$, we can rewrite the integral as$$- 6 \int \frac { \sin 2 x } { u ^ { 2 } } d u = - 6 \int \frac { 2 \sin x \cos x } { u ^ { 2 } } d u = - 12 \int \frac { \sin x } { u ^ { 2 } } \cos x d x$$Now, let $v = \cos x$. Then $dv = - \sin x dx$, and the integral becomes$$- 12 \int \frac { \sin x } { u ^ { 2 } } \cos x d x = - 12 \int \frac { 1 } { u ^ { 2 } } v d v = - 6 \arctan u + C = - 6 \arctan \left( \sec ^ { 2 } x \right) + C$$

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