Question

\[\int\frac{3 + 2 \cos x + 4 \sin x}{2 \sin x + \cos x + 3} \text{ dx }\]

Answer 1

\[\text{ Let I }= \int\left( \frac{3 + 2 \cos x + 4 \sin x}{2 \sin x + \cos x + 3} \right)dx\]
\[\text{ Let 3 }+ 2 \cos x + 4 \sin x = A \left( 2 \sin x + \cos x + 3 \right) + B \left( 2 \cos x - \sin x \right) + C\]
\[ \Rightarrow 3 + 2 \cos x + 4 \sin x = \left( 2A - B \right) \sin x + \left( A + 2B \right) \cos x + 3A + C\]

Comparing the coefficients of like terms

\[2A - B = 4 . . . \left( 1 \right)\]
\[A + 2B = 2 . . . (2)\]
\[3A + C = 3 . . . (3)\]

Multiplying eq (1) by 2 and adding it to eq (2) we get ,

\[\Rightarrow 4A - 2B + A + 2B = 8 + 2\]
\[ \Rightarrow 5A = 10\]
\[ \Rightarrow A = 2\]

Putting value of A = 2 in  eq (1)

\[\Rightarrow 2 \times 2 - B = 4\]
\[ \Rightarrow B = 0\]
\[\text{ Putting  value of   A   in eq (3) }\]
\[ \Rightarrow 3 \times 2 + C = 3\]
\[ \Rightarrow C = - 3\]

\[\therefore I = ∫ \left[ \frac{2 \left( 2 \sin x + \cos x + 3 \right) - 3}{2 \sin x + \cos x + 3} \right]dx\]
\[ = 2\ ∫   dx - 3\ ∫ \frac{1}{2 \sin x + \cos x + 3}dx\]
\[\text{ Substituting sin x }= \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} \text{ and }\cos x = \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}}\]
\[ \therefore I = 2\ ∫   dx - 3\ ∫  \frac{1}{2 \times \frac{2 \tan \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} + \frac{1 - \tan^2 \frac{x}{2}}{1 + \tan^2 \frac{x}{2}} + 3}dx\]
\[ = 2\  ∫   dx - 3\ ∫  \frac{\left( 1 + \tan^2 \frac{x}{2} \right)}{4 \tan \left( \frac{x}{2} \right) + 1 - \tan^2 \left( \frac{x}{2} \right) + 3 \left( 1 + \tan^2 \frac{x}{2} \right)}dx\]
\[ = 2\  ∫   dx - 3\ ∫ \frac{\sec^2 \left( \frac{x}{2} \right)}{2 \tan^2 \left( \frac{x}{2} \right) + 4 \tan \left( \frac{x}{2} \right) + 4} dx\]
\[ = 2\ ∫   dx - \frac{3}{2}\ ∫ \frac{\sec^2 \left( \frac{x}{2} \right)}{\tan^2 \left( \frac{x}{2} \right) + 2 \tan \left( \frac{x}{2} \right) + 2}dx\]
\[\text{  Putting tan } \left( \frac{x}{2} \right) = t\]
\[ \Rightarrow \frac{1}{2} \sec^2 \left( \frac{x}{2} \right) dx = dt\]
\[ \Rightarrow \sec^2 \left( \frac{x}{2} \right) dx = 2dt\]
\[ \therefore I = 2\ ∫  dx - \frac{3}{2}\ ∫ \frac{2}{t^2 + 2t + 2} dt\]
\[ = 2\ ∫   dx - 3\ ∫  \frac{1}{t^2 + 2t + 1 + 1}dt\]
\[ = 2\ ∫   dx - 3\ ∫  \frac{1}{\left( t + 1 \right)^2 + \left( 1 \right)^2}dt\]
\[ = 2x - \frac{3}{1} \tan^{- 1} \left( \frac{t + 1}{1} \right) + C\]
\[ = 2x - 3 \tan^{- 1} \left( \tan \frac{x}{2} + 1 \right) + C \left[ \because t = \tan \frac{x}{2} \right]\]