Evaluate the integral.
-$$\int _ { 0 } ^ { \pi / 8 } \left( 1 + e ^ { \tan 2 x } \right) \sec ^ { 2 } 2 x d x$$
A) $2 \mathrm { e }$
B) $- \frac { e } { 2 }$
C) $\frac { e } { 2 }$
D) e

Question

Quizplus · Accepted Answer

First, we use the substitution $u = 	an 2x$ and $du = 2\sec^2 2x dx$, which turns the integral into
$$\frac{1}{2} \int_0^{1} (1+e^u) du$$
Using the bounds of integration, this becomes
$$\frac{1}{2} \int_0^{	an(\pi/4)} (1+e^u) du = \frac{1}{2} \left[ u + e^u ight]_0^{	an(\pi/4)}$$
$$= \frac{1}{2} \left[ 	an 2x + e^{	an 2x} ight]_0^{\pi/8}$$
Substituting the limits of integration, this becomes
$$\frac{1}{2} \left[ 	an \frac{\pi}{4} - 	an 0 + e^{	an 0} - e^{-	an{\pi}/{4}} ight]$$
Simplifying this expression, we get
$$\frac{1}{2} \left[1 + 1 - \frac{1}{e^{1/2}} ight] = \frac{e}{2} - \frac{1}{2e^{1/2}}$$
Since $e^{1/2} > 1$ then $1/e^{1/2} < 1$ and so $\frac{1}{2e^{1/2}} < \frac{1}{2}$. Therefore, $\frac{e}{2} - \frac{1}{2e^{1/2}} < \frac{e}{2}$. So, the answer is $\boxed{	extbf{(C)}}$.

Evaluate the Integral ∫0π/8(1+etan⁡2x)sec⁡22xdx\int _ { 0 } ^ { \pi / 8 } \left( 1 + e ^ { \tan 2 x } \right) \sec ^ { 2 } 2 x d x∫0π/8​(1+etan2x)sec22xdx

Evaluate the Integral $\int _ { 0 } ^ { \pi / 8 } \left( 1 + e ^ { \tan 2 x } \right) \sec ^ { 2 } 2 x d x$