Question

$\int \frac{\left(3\sin x-2\right)\cos x}{5-{\cos }^{2}x-4\sin x}dx$

Answer 1

$$\begin{gathered} \int \frac{\left(3\sin x-2\right)\cos xdx}{5-{\cos }^{2}x-4\sin x} \\ =\int \frac{\left(3\sin x-2\right)\cos xdx}{5-\left(1-{\sin }^{2}x\right)-4\sin x} \\ =\int \frac{\left(3\sin x-2\right)\cos xdx}{{\sin }^{2}x-4\sin x+4} \\ \mathrm{Let}\sin x=t \\ \Rightarrow \cos xdx=dt \\ \int \frac{\left(3t-2\right)dt}{t^2-4t+4} \\ 3t-2=\mathrm{A}\frac{d}{dx}\left(t^2-4t+4\right)+\mathrm{B} \\ 3t-2=\mathrm{A}\left(2t-4\right)+\mathrm{B} \\ 3t-2=\left(2\mathrm{A}\right)t+\mathrm{B}-4\mathrm{A} \end{gathered}$$

Comparing the Coefficients of like powers of t

$$\begin{gathered} 2\mathrm{A}=3 \\ \mathrm{A}=\frac{3}{2} \\ \mathrm{B}-4\mathrm{A}=-2 \\ \mathrm{B}-4\times \frac{3}{2}=-2 \\ \mathrm{B}=-2+6 \\ \mathrm{B}=4 \end{gathered}$$

$$\begin{gathered} 3t-2=\frac{3}{2}\left(2t-4\right)+4 \\ \therefore \int \frac{\left(3t-2\right)dt}{t^2-4t+4} \\ =\int \left(\frac{{\displaystyle \frac{3}{2}}\left(2t-4\right)+4}{t^2-4t+4}\right)dt \\ =\frac{3}{2}\int \left(\frac{2t-4}{t^2-4t+4}\right)dt+4\int \frac{dt}{t^2-4t+4} \\ =\frac{3}{2}{\mathrm{I}}_1+4{\mathrm{I}}_2...\left(1\right) \\ \mathrm{where} \\ {\mathrm{I}}_1=\int \frac{\left(2t-4\right)dt}{t^2-4t+4},{\mathrm{I}}_2=\int \frac{dt}{t^2-4t+4} \\ {\mathrm{I}}_1=\int \frac{\left(2t-4\right)dt}{t^2-4t+4} \\ \mathrm{Let}{t}^2-4t+4=p \\ \Rightarrow \left(2t-4\right)dt=dp \\ {\mathrm{I}}_1=\int \frac{\left(2t-4\right)dt}{t^2-4t+4} \\ =\int \frac{dp}{p} \\ =\log \left|p\right|+C_1 \\ =\log \left|t^2-4t+4\right|+C_1...\left(2\right) \\ {\mathrm{I}}_2=\int \frac{dt}{t^2-4t+4} \\ {\mathrm{I}}_2=\int \frac{dt}{{\left(t-2\right)}^2} \\ {\mathrm{I}}_2=\int {\left(t-2\right)}^{-2}dt \\ {\mathrm{I}}_2=\frac{{\left(t-2\right)}^{-2+1}}{-2+1}+C_2 \\ {\mathrm{I}}_2=\frac{-1}{t-2}+C_2...\left(3\right) \\ \mathrm{from}\left(1\right),\left(2\right)\mathrm{and}\left(3\right) \\ \int \frac{\left(3\sin x-2\right)\cos xdx}{5-{\cos }^{2}x-4\sin x} \\ =\frac{3}{2}\log \left|t^2-4t+4\right|+4\times \frac{-1}{t-2}+C_1+C_2 \\ =\frac{3}{2}\log \left|{\sin }^{2}x-4\sin x+4\right|+\frac{4}{2-t}+C\left(\mathrm{Where}C=C_1+C_2\right) \\ =\frac{3}{2}\log \left|{\left(\sin x-2\right)}^2\right|+\frac{4}{2-\sin x}+C \\ =\frac{3}{2}\times 2\log \left|\sin x-2\right|+\frac{4}{2-\sin x}+C \\ =3\log \left|2-\sin x\right|+\frac{4}{2-\sin x}+C \end{gathered}$$