$\text { Find the integral } \int \frac { \sin x } { 13 + \cos ^ { 2 } x } d x \text {. }$
A) $- \frac { 1 } { 13 } \arctan ( \cos x \sqrt { 13 } ) + C$
B)$- \sqrt { 13 } \arctan \left( \frac { \sqrt { 13 } } { 13 } \cos x \right) + C$
C)$\sqrt { 13 } \arctan \left( \frac { \sqrt { 13 } } { 13 } \sin x \right) + C$
D) $\frac { \sqrt { 13 } } { 13 } \arctan \left( \frac { \sqrt { 13 } } { 13 } \sin x \right) + C$
E) $- \frac { \sqrt { 13 } } { 13 } \arctan \left( \frac { \sqrt { 13 } } { 13 } \cos x \right) + C$

Question

$\text { Find the integral } \int \frac { \sin x } { 13 + \cos ^ { 2 } x } d x \text {. }$ 
A) $- \frac { 1 } { 13 } \arctan ( \cos x \sqrt { 13 } ) + C$
B)$- \sqrt { 13 } \arctan \left( \frac { \sqrt { 13 } } { 13 } \cos x \right) + C$
C)$\sqrt { 13 } \arctan \left( \frac { \sqrt { 13 } } { 13 } \sin x \right) + C$
D) $\frac { \sqrt { 13 } } { 13 } \arctan \left( \frac { \sqrt { 13 } } { 13 } \sin x \right) + C$
E) $- \frac { \sqrt { 13 } } { 13 } \arctan \left( \frac { \sqrt { 13 } } { 13 } \cos x \right) + C$

Quizplus · Accepted Answer

The integral can be solved by recognizing that the derivative of $\cos x$ is $-\sin x$, which matches the numerator after considering a negative sign. The integral then becomes a standard form for the inverse tangent function, $\arctan$, with a coefficient adjustment to match the denominator's form. Option E correctly represents this solution, incorporating the negative sign and the correct argument for the $\arctan$ function.

Find the integral ∫sin⁡x13+cos⁡2xdx. \text { Find the integral } \int \frac { \sin x } { 13 + \cos ^ { 2 } x } d x \text {. } Find the integral ∫13+cos2xsinx​dx.

$\text { Find the integral } \int \frac { \sin x } { 13 + \cos ^ { 2 } x } d x \text {. }$