Question

\[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\]

Answer 1

I= \[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 - \cos^2 x - 7\sin x}dx\]

 

  =  \[\int\frac{\left( 3\sin x - 2 \right)\cos x}{13 -(1 -  \ sin ^2  x) - 7\sin x}dx\]    `(∵  cos^2x =1 - sin^2 x)`
\[ = \int\frac{\left( 3\sin x - 2 \right) \cos x}{\text{ sin^}2 x - 7\sin x + 12}dx\]
\[ = \int\frac{\left( 3\sin x - 2 \right) \cos x}{\text{ sin^}2 x - 4\sin x - 3\text{ sin } x + 12}dx\]
\[ = \int\frac{\left( 3\sin x - 2 \right) \cos x}{\sin x\left( \sin x - 4 \right) - 3\left( \sin x - 4 \right)}dx\]
\[ = \int\frac{\left( 3\sin x - 2 \right)\cos x}{\left( \sin x - 3 \right)\left( \sin x - 4 \right)}dx\]

\[\text{ Let sin x }= t\]
\[ \Rightarrow \text{  cos x dx }= dt\]
\[ \therefore I = \int\frac{\left( 3t - 2 \right)}{\left( t - 3 \right)\left( t - 4 \right)}dt\]

Using partial fraction, we get

\[\frac{\left( 3t - 2 \right)}{\left( t - 3 \right)\left( t - 4 \right)} = \frac{A}{\left( t - 3 \right)} + \frac{B}{\left( t - 4 \right)} = \frac{A\left( t - 4 \right) + B\left( t - 3 \right)}{\left( t - 3 \right)\left( t - 4 \right)}\]
\[ \Rightarrow 3t - 2 = (A + B)t - 4A - 3B\]

Comparing coefficients, we get

A = - 7 and B = 10

So, 

\[I = - 7\int\frac{1}{\left( t - 3 \right)}dt + 10\int\frac{1}{\left( t - 4 \right)}dt\]

\[\Rightarrow I = - 7\text{ ln }\left| t - 3 \right| + 10\text{ ln}\left| t - 4 \right| + c\]
\[ \therefore I = - 7\text{ ln }\left| \sin x - 3 \right| + 10 \text{ ln }\left| \sin x - 4 \right| + c\]