If I = $\int $C $( \sin 2 x - \tan y ) d x + x \sec ^ { 2 } y d y$ is independent of the path where C is a curve from $\left( 0 , \frac { \pi } { 4 } \right)$ to $\left( \frac { \pi } { 4 } , - \frac { \pi } { 4 } \right)$ then I is

A) $\frac { \pi + 2 } { 2 }$
B) $\frac { \pi - 2 } { 2 }$
C) $- \frac { \pi - 2 } { 4 }$
D) $\frac { \pi - 2 } { 4 }$
E) $\frac { \pi + 2 } { 4 }$

Question

If I = $\int $C $( \sin 2 x - \tan y ) d x + x \sec ^ { 2 } y d y$ is independent of the path where C is a curve from $\left( 0 , \frac { \pi } { 4 } \right)$ to $\left( \frac { \pi } { 4 } , - \frac { \pi } { 4 } \right)$ then I is

A) $\frac { \pi + 2 } { 2 }$
B) $\frac { \pi - 2 } { 2 }$
C) $- \frac { \pi - 2 } { 4 }$
D) $\frac { \pi - 2 } { 4 }$
E) $\frac { \pi + 2 } { 4 }$

Quizplus · Accepted Answer

The Answer of If I = $\int $C $( \sin 2 x - \tan y ) d x + x \sec ^ { 2 } y d y$ is independent of the path where C is a curve from $\left( 0 , \frac { \pi } { 4 } \right)$ to $\left( \frac { \pi } { 4 } , - \frac { \pi } { 4 } \right)$ then I is

A) $\frac { \pi + 2 } { 2 }$
B) $\frac { \pi - 2 } { 2 }$
C) $- \frac { \pi - 2 } { 4 }$
D) $\frac { \pi - 2 } { 4 }$
E) $\frac { \pi + 2 } { 4 }$

If I = ∫\int ∫ C (sin⁡2x−tan⁡y)dx+xsec⁡2ydy( \sin 2 x - \tan y ) d x + x \sec ^ { 2 } y d y(sin2x−tany)dx+xsec2ydy

If I = $\int$ C $( \sin 2 x - \tan y ) d x + x \sec ^ { 2 } y d y$