Find the derivative.
-$\frac { \mathrm { d } } { \mathrm { dt } } \int _ { 0 } ^ { \sin t } \frac { 1 } { 1 - \mathrm { u } ^ { 2 } } \mathrm { du }$
A) $\frac { - \cos t } { 1 - \sin ^ { 2 } t }$
B) $\frac { 1 } { 1 - \sin ^ { 2 } t }$
C) $\frac { \cos t } { 1 - \sin ^ { 2 } t }$
D) $\frac { 1 } { \cos t \left( 1 - \sin ^ { 2 } t \right) }$

Question

Quizplus · Accepted Answer

Let's first find the antiderivative of the integrand:

$$\int \frac{1}{1-u^{2}} du = \frac{1}{2}\ln\left|\frac{1+u}{1-u}ight|$$

Using the Fundamental Theorem of Calculus, we have:

$$\frac { \mathrm { d } } { \mathrm { dt } } \int _ { 0 } ^ { \sin t } \frac { 1 } { 1 - \mathrm { u } ^ { 2 } } \mathrm { du } = \frac{d}{dt}\left[\frac{1}{2}\ln\left|\frac{1+\sin t}{1-\sin t}ight|ight]$$

Using the chain rule:

$$\frac{d}{dt}\left[\frac{1}{2}\ln\left|\frac{1+\sin t}{1-\sin t}ight|ight] = \frac{1}{2}\cdot\frac{1}{\frac{1+\sin t}{1-\sin t}}\cdot\frac{d}{dt}\left[\frac{1+\sin t}{1-\sin t}ight]$$

Simplifying:

$$\frac{1}{2}\cdot\frac{1-\sin t}{1+\sin t}\cdot\frac{(1+\cos t) - (1-\cos t)}{(1-\sin t)^2} = \frac{1}{2}\cdot\frac{2\cos t}{(1-\sin^{2}t)}=\frac{\cos t}{1-\sin^{2}t}$$

Therefore, the answer is $\boxed{	extbf{(C)}}$

Find the Derivative ddt∫0sin⁡t11−u2du\frac { \mathrm { d } } { \mathrm { dt } } \int _ { 0 } ^ { \sin t } \frac { 1 } { 1 - \mathrm { u } ^ { 2 } } \mathrm { du }dtd​∫0sint​1−u21​du

Find the Derivative $\frac { \mathrm { d } } { \mathrm { dt } } \int _ { 0 } ^ { \sin t } \frac { 1 } { 1 - \mathrm { u } ^ { 2 } } \mathrm { du }$