Question

$\int \frac{1}{\cos x\left(5-4\sin x\right)}dx$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ I=\int \frac{dx}{\cos x\left(5-4\sin x\right)} \\ =\int \frac{\cos xdx}{{\cos }^{2}x\left(5-4\sin x\right)} \\ =\int \frac{\cos xdx}{\left(1-{\sin }^{2}x\right)\left(5-4\sin x\right)} \\ =\int \frac{\cos xdx}{\left(1-\sin x\right)\left(1+\sin x\right)\left(5-4\sin x\right)} \\ \mathrm{Putting}\sin x=t \\ \Rightarrow \cos xdx=dt \\ \therefore I=\int \frac{dt}{\left(1-t\right)\left(1+t\right)\left(5-4t\right)} \\ \mathrm{Let}\frac{1}{\left(1-t\right)\left(1+t\right)\left(5-4t\right)}=\frac{A}{1-t}+\frac{B}{1+t}+\frac{C}{5-4t} \\ \Rightarrow \frac{1}{\left(1-t\right)\left(1+t\right)\left(5-4t\right)}=\frac{A\left(1+t\right)\left(5-4t\right)+B\left(1-t\right)\left(5-4t\right)+C\left(1-t\right)\left(1+t\right)}{\left(1-t\right)\left(1+t\right)\left(5-4t\right)} \\ \Rightarrow 1=A\left(1+t\right)\left(5-4t\right)+B\left(1-t\right)\left(5-4t\right)+C\left(1-t\right)\left(1+t\right) \\ \mathrm{Putting}1+t=0 \\ \Rightarrow t=-1 \\ 1=B\left(2\right)\left(5+4\right) \\ B=\frac{1}{18} \\ \mathrm{Putting}1-t=0 \\ \Rightarrow t=1 \\ 1=A\left(2\right)\left(5-4\right)+B\times 0+C\times 0 \\ A=\frac{1}{2} \\ \mathrm{Putting}5-4t=0 \\ \Rightarrow 4t=5 \\ \Rightarrow t=\frac{5}{4} \\ 1=C\left(1-\frac{5}{4}\right)\left(1+\frac{5}{4}\right) \\ \Rightarrow 1=C\left(-\frac{1}{4}\right)\left(\frac{9}{4}\right) \\ \Rightarrow C=-\frac{16}{9} \\ \therefore I=\frac{1}{2}\int \frac{dt}{1-t}+\frac{1}{18}\int \frac{dt}{1+t}-\frac{16}{9}\int \frac{dt}{5-4t} \\ =\frac{1}{2}\frac{\log \left|1-t\right|}{-1}+\frac{1}{18}\log \left|1+t\right|-\frac{16}{9}\times \frac{\log \left|5-4t\right|}{-4}+C \\ =\frac{1}{18}\log \left|1+t\right|-\frac{1}{2}\log \left|1-t\right|+\frac{4}{9}\log \left|5-4t\right|+C \\ =\frac{1}{18}\log \left|1+\sin x\right|-\frac{1}{2}\log \left|1-\sin x\right|+\frac{4}{9}\log \left|5-4\sin x\right|+C \end{gathered}$$