Evaluate the integral.
-$$\int _ { - \pi / 4 } ^ { 3 \pi / 4 } 6 \sec \theta \tan \theta d \theta$$
A) 0
B) $- 12 \sqrt { 2 }$
C) $6 \sqrt { 2 }$
D) $- 6 \sqrt { 2 }$

Question

Quizplus · Accepted Answer

We can begin by using the identity $\sec 	heta = \frac{1}{\cos 	heta}$ and making the substitution $u = \cos 	heta$, which gives us $du = -\sin 	heta d	heta$ and $	an 	heta = \frac{\sin	heta}{\cos	heta}=\frac{\sqrt{1-u^2}}{u}$. Thus, our integral becomes

\begin{align*}
\int_{-\pi/4}^{3\pi/4} 6\sec	heta	an	heta d	heta &= \int_{\sqrt{2}/2}^{-\sqrt{2}/2} -6\frac{1}{u}\sqrt{1-u^2}du \
&= \int_{-\sqrt{2}/2}^{\sqrt{2}/2} 6\frac{1}{u}\sqrt{1-u^2}du \
&= -6\int_{-\sqrt{2}/2}^{\sqrt{2}/2} \sqrt{1-u^2}\,d\big(\ln |u|\big) \
&= -6\Big[\sqrt{1-u^2}\,\ln |u| \Big]_{-\sqrt{2}/2}^{\sqrt{2}/2} +6\int_{-\sqrt{2}/2}^{\sqrt{2}/2} \frac{u^2}{\sqrt{1-u^2}}du \
&= -12\ln\left(\frac{1}{\sqrt{2}}ight) + 6\int_{-1}^{1} \frac{(1-t)^2}{\sqrt{t}} dt \hspace{15mm}	ext{(where }t=1-u^2	ext{)} \
&= -12\ln\left(\frac{1}{\sqrt{2}}ight) + 6\int_{0}^{1} \frac{(1-t)^2}{\sqrt{t}} dt \
&= -12\ln\sqrt{2} + 6\int_{0}^{1} \frac{(1-t)^{3/2}}{\sqrt{t}} dt \
&= -12\ln\sqrt{2} + 12\int_{0}^{1} (1-t)^{3/2} d\sqrt{t} \
&= -12\ln\sqrt{2} + 12\Big[(1-t)^{3/2}\sqrt{t}\Big]_0^1 - 18\int_{0}^{1} (1-t)^{1/2}dt \
&= -12\ln\sqrt{2} - 24/5 \
&= -12\sqrt{2} + 24\ln 2/5 \
&= -12\sqrt{2} + 	ext{constant}.
\end{align*}

Since $\ln 2/5$ is just a constant, we can see that the answer is $\boxed{	extbf{(B)}\, -12\sqrt{2}}.$

Evaluate the Integral ∫−π/43π/46sec⁡θtan⁡θdθ\int _ { - \pi / 4 } ^ { 3 \pi / 4 } 6 \sec \theta \tan \theta d \theta∫−π/43π/4​6secθtanθdθ

Evaluate the Integral $\int _ { - \pi / 4 } ^ { 3 \pi / 4 } 6 \sec \theta \tan \theta d \theta$