Question

\[\int\frac{5 \cos x + 6}{2 \cos x + \sin x + 3} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int\left( \frac{5 \cos x + 6}{2 \cos x + \sin x + 3} \right)dx\]
\[\text{ and  let 5 cos x + 6 }= A \left( 2 \ cosx + \sin x + 3 \right) + B\left( - 2 \sin x + \cos x \right) + C . . . . (1) \]
\[ \Rightarrow 5 \cos x + 6 = \left( A - 2B \right) \sin x + \left( 2A + B \right) \cos x + 3A + C\]

Comparing coefficients of like terms

\[A - 2B = 0 . . . \left( 2 \right)\]
\[2A + B = 5 . . . (3)\]
\[3A + C = 6 . . . (4)\]

Multiplying eq (3) by 2 and then adding to eq (2)

4A + 2B + A – 2B = 10

\[\Rightarrow\]A = 2

Putting value of A in eq (2) and eq (4) we get,
B = 1& C = 0

\[\text{ By putting the values of A, B and C in eq (1) we get ,} \]
\[ \therefore I = \int\left[ \frac{2 \left( 2 \cos x + \sin x + 3 \right) + \left( - 2 \sin x + \cos x \right)}{\left( 2 \cos x + \sin x + 3 \right)} \right]dx\]
\[ = 2\int dx + \int \left( \frac{- 2 \sin x + \cos x}{2 \cos x + \sin x + 3} \right)dx\]
\[\text{ Putting 2 cos x + sin x + 3 = t }\]
\[ \Rightarrow \left( - 2 \sin x + \cos x \right)dx = dt\]
\[ \therefore I = 2\int dx + \int\frac{1}{t}dt\]
\[ = 2x + \text{ ln }\left| 2 \cos x + \sin x + 3 \right| + C\]