Question

\[\int\frac{4 \sin x + 5 \cos x}{5 \sin x + 4 \cos x} \text{ dx }\]

Answer 1

\[\text{ Let I } = \int\left( \frac{4 \sin x + 5 \cos x}{5 \sin x + 4 \cos x} \right)dx\]
\[and\text{ let 4 sin x + 5 cos x = A } \left( 5 \sin x + 4 \cos x \right) + B \left( 5 \cos x - 4 \sin x \right) . . . (1) \]
\[ \Rightarrow 4 \sin x + 5 \cos x = \left( 5A - 4B \right) \sin x + \left( 4A + 5B \right) \cos x\]
\[\text{By equating the coefficients of like terms we get}, \]
\[5A - 4B = 4 . . . \left( 2 \right)\]
\[4A + 5B = 5 . . . \left( 3 \right)\]

By solving eq (2) and eq (3) we get,

\[A = \frac{40}{41}, B = \frac{9}{41}\]
\[\text{Thus, by substituting the values of A and B in eq} (1) , we get\]
\[I = \int\left[ \frac{\frac{40}{41}\left( 5 \ sinx + 4 \cos x \right) + \frac{9}{41}\left( 5 \ cos x - 4 \ sin x \right)}{\left( 5 \sin x + 4 \cos x \right)} \right]dx\]
\[ = \frac{40}{41}\int dx + \frac{9}{41}\int\left( \frac{5 \cos x - 4 \sin x}{5 \sin x + 4 \cos x} \right)dx\]
\[\text{ Putting 5 sin x + 4 cos x = t}\]
\[ \Rightarrow \left( 5 \cos x - 4 \sin x \right)dx = dt\]
\[ \therefore I = \frac{40}{41}x + \frac{9}{41}\int\frac{1}{t}dt\]
\[ = \frac{40}{41}x + \frac{9}{41} \text{ ln }\left| t \right| + C\]
\[ = \frac{40}{41}x + \frac{9}{41} \text{ ln } \left| 5 \sin x + 4 \cos x \right| + C\]