Evaluate the integral by making a substitution (possibly trigonometric) and then applying a reduction formula.
-$$\int e ^ { t } \sec ^ { 3 } \left( e ^ { t } - 2 \right) d t$$
A) $- \frac { 1 } { 2 } \left[ \sec \left( e ^ { t } - 2 \right) \tan \left( e ^ { t } - 2 \right) - \ln \left| \sec \left( e ^ { t } - 2 \right) + \tan \left( e ^ { t } - 2 \right) \right| \right] + C$
B) $\frac { 1 } { 2 } \left[ \sec \left( e ^ { t } - 2 \right) \tan \left( e ^ { t } - 2 \right) - \ln \left| \sec \left( e ^ { t } - 2 \right) + \tan \left( e ^ { t } - 2 \right) \right| \right] + C$
C) $- \frac { 1 } { 2 } \left[ \sec \left( e ^ { t } - 2 \right) \tan \left( e ^ { t } - 2 \right) + \ln \left| \sec \left( e ^ { t } - 2 \right) + \tan \left( e ^ { t } - 2 \right) \right| \right] + C$
D) $\frac { 1 } { 2 } \left[ \sec \left( e ^ { t } - 2 \right) \tan \left( e ^ { t } - 2 \right) + \ln \left| \sec \left( e ^ { t } - 2 \right) + \tan \left( e ^ { t } - 2 \right) \right| \right] + C$

Question

Quizplus · Accepted Answer

Let $u=e^t-2$. Then $du=e^t\ dt$ and $\sec^3(u+2)=\sec(u+2)\cdot\sec^2(u+2)=\sec(u+2)\cdot(\tan^2(u+2)+1)$. Therefore,
$$
\begin{aligned}
\int e^t \sec^3(e^t-2)\ dt &=\int \frac{1}{e^t}(u+2)(\tan^2u+1)\ du \
&= \int (u+2)\left(1+\tan^2u\right)\frac{1}{e^t}\ du\
&= \int (u+2)\sec^2u\frac{1}{e^t}\ du\
&= \frac{1}{e^t}\int (u+2)\sec^2u\ du\
&= \frac{1}{e^t}\left(\frac{1}{2}(u+2)\sec u\tan u+\frac{1}{2}\int\sec u\tan u\ du\right)\
&= \frac{1}{e^t}\left(\frac{1}{2}(e^t-2+2)\sec(e^t-2)\tan(e^t-2)-\frac{1}{2}\ln|\sec(e^t-2)+\tan(e^t-2)|\right)\
&= \frac{1}{2}\left[\sec(e^t-2)\tan(e^t-2)-\frac{1}{e^t}\ln|\sec(e^t-2)+\tan(e^t-2)|\right]+C.
\end{aligned}
$$
Therefore, the best choice is $\boxed{\text{(D)}}.$

Evaluate the Integral by Making a Substitution (Possibly Trigonometric) and Then