Question
Evaluate the following intregals:
$\int\frac{\cos\text{x}}{(1-\sin\text{x})^3(2+\sin\text{x})}\ \text{dx}$
$\int\frac{\cos\text{x}}{(1-\sin\text{x})^3(2+\sin\text{x})}\ \text{dx}$
✓
Answer
Let
$\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}=\text{dt}$
$\therefore\int\frac{\cos\text{x}}{(1+\sin\text{x})^3(2+\sin\text{x})}=\int\frac{1}{(1-\text{t})^3(2+\text{t})}\ \text{dt}$
Let $\text{f}(\text{t})=\frac{1}{(1-\text{t})^3(2+\text{t})}$
Then suppose
$\frac{1}{(1-\text{t})^3(2+\text{t})}=\frac{\text{A}}{1-\text{t}}+\frac{\text{B}}{(1-\text{t})^2}+\frac{\text{C}}{(1-\text{t})^3}+\frac{\text{D}}{(2+\text{t})}$
$\Rightarrow1=\text{A}(1-\text{t})^2(2+\text{t})+\text{B}(1-\text{t})(2+\text{t})\\+\text{C}(2+\text{t})+\text{D}(1-\text{t})^3$
Put t = 1
1 = 27D
$\Rightarrow\text{D}=\frac{1}{27}$
Similarly, we can find that $\text{A}=\frac{-1}{27}$ and $\text{B}=\frac{+1}{9}$
$\therefore\int\frac{1}{(1-\text{t})^3(2+\text{t})}\ \text{dt}=\frac{-1}{27}\int\frac{1}{1-\text{t}}\ \text{dt}+\frac{1}{9}\int\frac{\text{dt}}{(1-\text{t})^2}\\+\frac{1}{3}\int\frac{\text{dt}}{(1-\text{t})^3}+\frac{1}{27}\int\frac{\text{dt}}{2+\text{t}}$
$=\frac{-1}{27}\log|1-\text{t}|+\frac{1}{9(1-\text{t})}+\frac{1}{6(1-\text{t})^2}+\frac{1}{27}\log|2+\text{t}|+\text{C}$
Putting $\text{t}=\sin\text{x}$ we get
$\int\frac{\cos\text{x}}{(1-\sin\text{x})^3(2+\sin\text{x})}\ \text{dx}$
$=\frac{-1}{27}\log|1-\sin\text{x}|+\frac{1}{9(1-\sin\text{x})}\\+\frac{1}{6(1-\sin\text{x})^2}+\frac{1}{27}\log|2+\sin\text{x}|+\text{C}$
$\sin\text{x}=\text{t}$
$\Rightarrow\cos\text{x}=\text{dt}$
$\therefore\int\frac{\cos\text{x}}{(1+\sin\text{x})^3(2+\sin\text{x})}=\int\frac{1}{(1-\text{t})^3(2+\text{t})}\ \text{dt}$
Let $\text{f}(\text{t})=\frac{1}{(1-\text{t})^3(2+\text{t})}$
Then suppose
$\frac{1}{(1-\text{t})^3(2+\text{t})}=\frac{\text{A}}{1-\text{t}}+\frac{\text{B}}{(1-\text{t})^2}+\frac{\text{C}}{(1-\text{t})^3}+\frac{\text{D}}{(2+\text{t})}$
$\Rightarrow1=\text{A}(1-\text{t})^2(2+\text{t})+\text{B}(1-\text{t})(2+\text{t})\\+\text{C}(2+\text{t})+\text{D}(1-\text{t})^3$
Put t = 1
1 = 27D
$\Rightarrow\text{D}=\frac{1}{27}$
Similarly, we can find that $\text{A}=\frac{-1}{27}$ and $\text{B}=\frac{+1}{9}$
$\therefore\int\frac{1}{(1-\text{t})^3(2+\text{t})}\ \text{dt}=\frac{-1}{27}\int\frac{1}{1-\text{t}}\ \text{dt}+\frac{1}{9}\int\frac{\text{dt}}{(1-\text{t})^2}\\+\frac{1}{3}\int\frac{\text{dt}}{(1-\text{t})^3}+\frac{1}{27}\int\frac{\text{dt}}{2+\text{t}}$
$=\frac{-1}{27}\log|1-\text{t}|+\frac{1}{9(1-\text{t})}+\frac{1}{6(1-\text{t})^2}+\frac{1}{27}\log|2+\text{t}|+\text{C}$
Putting $\text{t}=\sin\text{x}$ we get
$\int\frac{\cos\text{x}}{(1-\sin\text{x})^3(2+\sin\text{x})}\ \text{dx}$
$=\frac{-1}{27}\log|1-\sin\text{x}|+\frac{1}{9(1-\sin\text{x})}\\+\frac{1}{6(1-\sin\text{x})^2}+\frac{1}{27}\log|2+\sin\text{x}|+\text{C}$
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