Evaluate the integral.
-$$\int _ { 0 } ^ { 5 \pi / 4 } \tan \frac { x } { 5 } d x$$
A) $\frac { 5 \ln 2 } { 2 }$
B) $\frac { - 5 \sqrt { 2 } } { 2 }$
C) $\frac { - 5 \ln 2 } { 2 }$
D) $\frac { 5 \sqrt { 2 } } { 2 }$

Question

Quizplus · Accepted Answer

The integral of $\tan \frac{x}{5}$ can be solved by using the substitution method. Let $u = \frac{x}{5}$, which implies $du = \frac{1}{5}dx$, or $dx = 5du$. Substituting these into the integral gives $\int \tan(u) \cdot 5du$, which simplifies to $5 \int \tan(u) du$. The integral of $\tan(u)$ is $\ln|\sec(u)| + C$, so the integral becomes $5\ln|\sec(u)| + C$. Substituting back for $u = \frac{x}{5}$, we get $5\ln|\sec(\frac{x}{5})| + C$. Evaluating this from $0$ to $5\pi/4$ gives $5[\ln|\sec(5\pi/4/5)| - \ln|\sec(0)|]$, which simplifies to $5[\ln|\sec(\pi/4)| - \ln|\sec(0)|]$. Since $\sec(\pi/4) = \sqrt{2}$ and $\sec(0) = 1$, the result is $5[\ln|\sqrt{2}| - \ln|1|] = 5\ln\sqrt{2}$. Using the property of logarithms that $\ln a^b = b\ln a$, this becomes $5\frac{1}{2}\ln 2 = \frac{5\ln 2}{2}$.

Evaluate the Integral ∫05π/4tan⁡x5dx\int _ { 0 } ^ { 5 \pi / 4 } \tan \frac { x } { 5 } d x∫05π/4​tan5x​dx

Evaluate the Integral $\int _ { 0 } ^ { 5 \pi / 4 } \tan \frac { x } { 5 } d x$