Question

$\int \frac{3+2\cos x+4\sin x}{2\sin x+\cos x+3}dx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \left(\frac{3+2\cos x+4\sin x}{2\sin x+\cos x+3}\right)dx \\ \mathrm{Let}3+2\cos x+4\sin x=A\left(2\sin x+\cos x+3\right)+B\left(2\cos x-\sin x\right)+C \\ \Rightarrow 3+2\cos x+4\sin x=\left(2A-B\right)\sin x+\left(A+2B\right)\cos x+3A+C \end{gathered}$$

Comparing the coefficients of like terms
$$\begin{gathered} 2A-B=4...\left(1\right) \\ A+2B=2...\left(2\right) \\ 3A+C=3...\left(3\right) \end{gathered}$$

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


$$\begin{gathered} \Rightarrow 4A-2B+A+2B=8+2 \\ \Rightarrow 5A=10 \\ \Rightarrow A=2 \end{gathered}$$

Putting value of A = 2 in eq (1)

$$\begin{gathered} \Rightarrow 2\times 2-B=4 \\ \Rightarrow B=0 \\ \mathrm{Putting}\mathrm{value}\mathrm{of}A\mathrm{in}\mathrm{eq}\left(3\right) \\ \Rightarrow 3\times 2+C=3 \\ \Rightarrow C=-3 \end{gathered}$$

$$\begin{gathered} \therefore I=\int \left[\frac{2\left(2\sin x+\cos x+3\right)-3}{2\sin x+\cos x+3}\right]dx \\ =2\int dx-3\int \frac{1}{2\sin x+\cos x+3}dx \\ \mathrm{Substituting}\sin x=\frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}\mathrm{and}\cos x=\frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}} \\ \therefore I=2\int dx-3\int \frac{1}{2\times {\displaystyle \frac{2\tan {\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}+{\displaystyle \frac{1-{\tan }^2{\displaystyle \frac{x}{2}}}{1+{\tan }^2{\displaystyle \frac{x}{2}}}}+3}dx \\ =2\int dx-3\int \frac{\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)}{4\tan \left({\displaystyle \frac{x}{2}}\right)+1-{\tan }^2\left({\displaystyle \frac{x}{2}}\right)+3\left(1+{\tan }^2{\displaystyle \frac{x}{2}}\right)}dx \\ =2\int dx-3\int \frac{{\sec }^2\left({\displaystyle \frac{x}{2}}\right)}{2{\tan }^2\left({\displaystyle \frac{x}{2}}\right)+4\tan \left({\displaystyle \frac{x}{2}}\right)+4}dx \\ =2\int dx-\frac{3}{2}\int \frac{{\sec }^2\left({\displaystyle \frac{x}{2}}\right)}{{\tan }^2\left({\displaystyle \frac{x}{2}}\right)+2\tan \left({\displaystyle \frac{x}{2}}\right)+2}dx \\ \mathrm{Putting}\tan \left(\frac{x}{2}\right)=t \\ \Rightarrow \frac{1}{2}{\sec }^2\left(\frac{x}{2}\right)dx=dt \\ \Rightarrow {\sec }^2\left(\frac{x}{2}\right)dx=2dt \\ \therefore I=2\int dx-\frac{3}{2}\int \frac{2}{t^2+2t+2}dt \\ =2\int dx-3\int \frac{1}{t^2+2t+1+1}dt \\ =2\int dx-3\int \frac{1}{{\left(t+1\right)}^2+{\left(1\right)}^2}dt \\ =2x-\frac{3}{1}{\tan }^{-1}\left(\frac{t+1}{1}\right)+C \\ =2x-3{\tan }^{-1}\left(\tan \frac{x}{2}+1\right)+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because \mathit{}t=\tan \mathit{}\frac{x}{2}\right] \end{gathered}$$