By part 2 of the Fundamental Theorem of Calculus, $\int _ { \frac { \pi } { 4 } } ^ { \frac { \pi } { 3 } } \sec x \tan x d x$ is

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

Question

By part 2 of the Fundamental Theorem of Calculus, $\int _ { \frac { \pi } { 4 } } ^ { \frac { \pi } { 3 } } \sec x \tan x d x$ is

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

Quizplus · Accepted Answer

The integral of $\sec x \tan x$ is $\sec x$, so we evaluate $\sec x$ from $\frac{\pi}{4}$ to $\frac{\pi}{3}$ to get $\sec(\frac{\pi}{3}) - \sec(\frac{\pi}{4}) = 2 - \sqrt{2}$.

By Part 2 of the Fundamental Theorem of Calculus ∫π4π3sec⁡xtan⁡xdx\int _ { \frac { \pi } { 4 } } ^ { \frac { \pi } { 3 } } \sec x \tan x d x∫4π​3π​​secxtanxdx

By Part 2 of the Fundamental Theorem of Calculus $\int _ { \frac { \pi } { 4 } } ^ { \frac { \pi } { 3 } } \sec x \tan x d x$