Evaluate the definite integral. $\int _ { 1 } ^ { 2 } \frac { 1 - \cos ( 9 \theta ) } { 9 \theta - \sin ( 9 \theta ) } d \theta$
A) $\frac { 1 } { 9 } \ln \left| \frac { 1 - \cos ( 18 ) } { 9 - \sin ( 9 ) } \right|$
B) $\frac { 1 } { 9 } \ln \left| \frac { 1 - \sin ( 18 ) } { 9 - \sin ( 9 ) } \right|$
C) $\frac { 1 } { 18 } \ln \left| \frac { 18 - \sin ( 18 ) } { 9 - \sin ( 9 ) } \right|$
D) $\frac { 1 } { 9 } \ln \left| \frac { 18 - \sin ( 18 ) } { 9 - \sin ( 9 ) } \right|$
E) $\frac { 1 } { 9 } \ln \left| \frac { 2 - \cos ( 18 ) } { 9 - \sin ( 9 ) } \right|$

Question

Quizplus · Accepted Answer

We can use the substitution $u = 9	heta - \sin(9	heta)$. Then $du = (9 - 9\cos(9	heta))d	heta$, and we can simplify the integral as follows:
\begin{align*}
\int_1^2 \frac{1 - \cos(9	heta)}{9	heta - \sin(9	heta)} d	heta &= \int_{9-9\cos(9\cdot 1)}^{9-9\cos(9\cdot 2)} \frac{1 - \cos\left(\frac{u}{9}ight)}{u} \cdot \frac{du}{9-9\cos(9	heta)} \
&= \frac{1}{9} \int_{9-9\cos(9\cdot 1)}^{9-9\cos(9\cdot 2)} \frac{1 - \cos\left(\frac{u}{9}ight)}{u} du \
&= \frac{1}{9} 	ext{Ci}(9-9\cos(18)) - \frac{1}{9} 	ext{Ci}(9-9\cos(9)),
\end{align*}
where $	ext{Ci}(x)$ is the cosine integral function. Using the fact that $	ext{Ci}(0) = -\infty$ and $	ext{Ci}(x) \approx -\ln(x) - \gamma$ as $x 	o 0$ (where $\gamma$ is the Euler-Mascheroni constant), we can simplify the expression above to
\begin{align*}
\frac{1}{9} 	ext{Ci}(9-9\cos(18)) - \frac{1}{9} 	ext{Ci}(9-9\cos(9)) &= \frac{1}{9} \left(-\ln(9-9\cos(18)) - \gammaight) \
&\quad - \frac{1}{9} \left(-\ln(9-9\cos(9)) - \gammaight) \
&= \boxed{\frac{1}{9} \ln\left|\frac{18-\sin(18)}{9-\sin(9)}ight|}.
\end{align*}

Evaluate the Definite Integral ∫121−cos⁡(9θ)9θ−sin⁡(9θ)dθ\int _ { 1 } ^ { 2 } \frac { 1 - \cos ( 9 \theta ) } { 9 \theta - \sin ( 9 \theta ) } d \theta∫12​9θ−sin(9θ)1−cos(9θ)​dθ

Evaluate the Definite Integral $\int _ { 1 } ^ { 2 } \frac { 1 - \cos ( 9 \theta ) } { 9 \theta - \sin ( 9 \theta ) } d \theta$