Evaluate the integral by making a substitution and then using a table of integrals.
-$$\int \tan t \cdot \sqrt { 1 - \cos ^ { 2 } t } d t$$
A) $\ln | \csc t - \cot t | - \sin t + C$
B) $\ln | \sec t + \tan t | - \sin t + C$
C) $\ln | \sec t + \tan t | + \sin t + C$
D) $\ln | \sec t | + \cos t + C$

Question

Quizplus · Accepted Answer

Make the substitution $u=\cos t$, then $du=-\sin t\,dt$ and the integral becomes
$$\int \sqrt{1-u^2}\cdot\frac{	an^{-1}\sqrt{1-u^2}}{\sqrt{1-u^2}}\,du=\int\frac{	an^{-1}\sqrt{1-u^2}}{u^2}\,du.$$
Using the table of integrals, we have
$$\int\frac{	an^{-1}\sqrt{1-u^2}}{u^2}\,du=-\frac{\sqrt{1-u^2}\,	an^{-1}\sqrt{1-u^2}}{u}+\ln|u|+C.$$
Substituting back $u=\cos t$ and simplifying, we get
$$\int 	an t \cdot \sqrt { 1 - \cos ^ { 2 } t } d t = -\sqrt { 1 - \cos ^ { 2 } t } \cdot 	an ^ { - 1 } \sqrt { 1 - \cos ^ { 2 } t } + \ln | \cos t | + C.$$
Using trigonometric identity $	an^{-1}\sqrt{1-\cos^2 t}=\ln|\sec t+	an t|$, we obtain the answer
$$\int 	an t \cdot \sqrt { 1 - \cos ^ { 2 } t } d t = \ln | \sec t + 	an t | - \sin t + C.$$

Evaluate the Integral by Making a Substitution and Then Using ∫tan⁡t⋅1−cos⁡2tdt\int \tan t \cdot \sqrt { 1 - \cos ^ { 2 } t } d t∫tant⋅1−cos2t​dt

Evaluate the Integral by Making a Substitution and Then Using $\int \tan t \cdot \sqrt { 1 - \cos ^ { 2 } t } d t$