Evaluate the integral.
-$$\int \frac { \cos t d t } { \sin ^ { 2 } t - 4 \sin t + 3 }$$
A) $\frac { 1 } { 2 } \ln | \mathrm { t } - 3 | - \frac { 1 } { 2 } \ln | \mathrm { t } - 1 | + \mathrm { C }$
B) $\frac { 1 } { 2 } \ln | \sin t - 3 | + \frac { 1 } { 2 } \ln | \sin t - 1 | + C$
C) $\frac { 1 } { 2 } \ln | \sin t - 3 | - \frac { 1 } { 2 } \ln | \sin t - 1 | + C$
D) $\ln | \sin t - 3 | - \ln | \sin t - 1 | + C$

Question

Quizplus · Accepted Answer

We can start by factoring the denominator:
$$\sin^2t-4\sin t + 3 = (\sin t - 3)(\sin t - 1)$$
Then we can apply partial fractions:
$$\frac{\cos t}{\sin^2t-4\sin t + 3}=\frac{A}{\sin t-3}+\frac{B}{\sin t-1}$$
Multiplying both sides by the denominator yields:
$$\cos t = A(\sin t - 1) + B(\sin t - 3)$$
Substituting $\sin t=1$ and $\sin t=3$ gives us a system of equations for $A$ and $B$:
$$\begin{cases} A-B=1 \ 3B-A=1 \end{cases} \implies A=1,B=0$$
Thus we have:
$$\frac{\cos t}{\sin^2t-4\sin t + 3}=\frac{1}{\sin t-3}+\frac{0}{\sin t-1}$$
Integrating both sides gives:
$$\int \frac{\cos t}{\sin^2t-4\sin t + 3} dt= \ln|\sin t -3| +C$$
Therefore, the correct answer is (C).

Evaluate the Integral ∫cos⁡tdtsin⁡2t−4sin⁡t+3\int \frac { \cos t d t } { \sin ^ { 2 } t - 4 \sin t + 3 }∫sin2t−4sint+3costdt​

Evaluate the Integral $\int \frac { \cos t d t } { \sin ^ { 2 } t - 4 \sin t + 3 }$