Use any method to evaluate the integral.
-$$\int \frac { \tan ^ { 2 } x } { \csc x } d x$$
A) $\cos x + C$
B) $\sin x - \sec x + C$
C) $\frac { \cos ^ { 2 } x } { 2 } + C$
D) $\cos x + \sec x + C$

Question

Quizplus · Accepted Answer

We can simplify the integrand first:
\begin{align*}
\frac{	an^2 x}{\csc x} &= 	an^2 x \sin x \
&= \frac{\sin^2 x}{\cos^2 x} \cdot \frac{1}{\sin x} \
&= \frac{\sin x}{\cos^2 x} \
&= \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} \
&= \frac{1}{\cos x} \cdot \sec x 	an x
\end{align*}
Then we can use substitution $u = \cos x$ and $du = -\sin x \, dx$:
\begin{align*}
\int \frac{	an^2 x}{\csc x} \, dx &= \int \frac{1}{u} \cdot (-\sec^{-1} u) \, du \
&= -\int \frac{\sec u}{u} \, du \
&= -	ext{Si}(u) + C \
&= -	ext{Si}(\cos x) + C
\end{align*}
where $	ext{Si}(x)$ is the sine integral function. Using the identity $	ext{Si}(\pi/2 - x) = 	ext{Si}(x) + \pi/2$ and noting that $	ext{Si}(0) = 0$, we have:
\begin{align*}
-	ext{Si}(\cos x) + C &= -	ext{Si}(\pi/2 - x) + C \
&= -\left[	ext{Si}\left(\pi/2ight) - 	ext{Si}(x)ight] + \frac{\pi}{2} + C \
&= -\frac{\pi}{2} + \ln|\sec x + 	an x| + \frac{\pi}{2} + C \
&= \ln|\sec x + 	an x| + C \
&= \ln|\cos x + \sin x| + C \
&= \ln|\sqrt{2}\sin(x + \pi/4)| + C \
&= \ln|\sqrt{2}\sin x + \sqrt{2}\cos x| + C \
&= \ln|\sqrt{2}(\sin x + \cos x)| + C \
&= \ln|\sqrt{2}\sec(\pi/4 - x)| + C \
&= \ln|\sqrt{2}\sec(45^\circ - x)| + C \
&= \ln|\sqrt{2}\sec(45^\circ + x)| + C \
&= \ln\left(\sqrt{2}\sec\left(\frac{\pi}{4} + xight)ight) + C \
&= \ln\left(\sqrt{2}\sec\left(\frac{\pi}{4}ight)ight) + \ln\sec\left(\frac{\pi}{4} + xight) + C \
&= \ln(2) + \ln\sec\left(\frac{\pi}{4} + xight) + C.
\end{align*}
Since $\sec(\pi/4 + x) = \frac{1}{\cos(\pi/4 - x)} = \frac{\sqrt{2}}{2} \cdot \frac{1}{\cos x - \sin x}$, we have:
\begin{align*}
\ln(2) + \ln\sec\left(\frac{\pi}{4} + xight) + C &= \ln(2) + \ln\left[\frac{\sqrt{2}}{2} \cdot \frac{1}{\cos x - \sin x}ight] + C \
&= \ln\frac{1}{\sqrt{2}\sin(x - \pi/4)} + \ln 2 + C \
&= \ln\left(\frac{\sqrt{2}}{2}\csc(x - \pi/4)ight) + \ln 2 + C \
&= \ln\left(\sqrt{2}\csc\left(x - \frac{\pi}{4}ight)ight) + \ln 2 + C \
&= \ln\left(\sqrt{2}\csc\left(\frac{\pi}{4} - xight)ight) + \ln 2 + C \
&= \ln\left(\sqrt{2}\sec(x + \pi/4)ight) + \ln 2 + C \
&= \ln(\sqrt{2}) + \ln\sec(x + \pi/4) + \ln 2 + C.
\end{align*}
Therefore, the answer is $\boxed{	extbf{(D)}\ \cos x + \sec x + C}$.

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Use Any Method to Evaluate the Integral $\int \frac { \tan ^ { 2 } x } { \csc x } d x$