Question
Evaluate the following integrals:
$\int\frac{1}{\sin\text{x}(3+2\cos\text{x})}\ \text{dx}$
$\int\frac{1}{\sin\text{x}(3+2\cos\text{x})}\ \text{dx}$
✓
Answer
Let $\text{I}=\int\frac{1}{\sin\text{x}(3+2\cos\text{x})}\ \text{dx}$
$=\frac{\sin\text{x dx}}{\sin^2\text{x}(3+2\cos\text{x })}$
$=\frac{\sin\text{x dx}}{(1-\cos^2\text{x})(3+2\text{x})}$
Let $\cos\text{x}=\text{t}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{(\text{t}^2-1)(3+2\text{t})}$
Now,
Let $\frac{1}{(\text{t}^2-1)(3+2\text{t})}=\frac{\text{A}}{\text{t}-1}+\frac{\text{B}}{\text{t}+1}+\frac{\text{C}}{3+2\text{t}}$
$\Rightarrow 1 = A(t + 1)(3 + 2t) + B(t - 1)(3 + 2t) + C (t^2- 1)$
Put $t = -1$
$\Rightarrow 1 = -2B$ $\Rightarrow\text{A}=\frac{1}{10}$
Put $\text{t}=-\frac{3}{2}$
$\Rightarrow1=\frac{5}{4}\text{C}\Rightarrow\text{C}=\frac{4}{5}$
Thus,
$\text{I}=\frac{1}{10}\int\frac{\text{dt}}{\text{t}-1}-\frac{1}{2}\int\frac{\text{dt}}{\text{t}+1}+\frac{5}{4}\int\frac{\text{dt}}{3+2\text{t}}$
$=\frac{1}{10}\log|\text{t}-1|=\frac{1}{2}\log|\text{t}+1|+\frac{2}{5}\log|3+2\text{t}|+\text{C}$
Hence,
$\text{I}=\frac{1}{10}\log|\cos\text{x}-1|-\frac{1}{2}\log|\cos\text{x}+1|+\frac{2}{5}\log|3+2\cos\text{x}|+\text{C}$
$=\frac{\sin\text{x dx}}{\sin^2\text{x}(3+2\cos\text{x })}$
$=\frac{\sin\text{x dx}}{(1-\cos^2\text{x})(3+2\text{x})}$
Let $\cos\text{x}=\text{t}$
$\Rightarrow-\sin\text{x dx}=\text{dt}$
$\therefore\text{I}=\int\frac{\text{dt}}{(\text{t}^2-1)(3+2\text{t})}$
Now,
Let $\frac{1}{(\text{t}^2-1)(3+2\text{t})}=\frac{\text{A}}{\text{t}-1}+\frac{\text{B}}{\text{t}+1}+\frac{\text{C}}{3+2\text{t}}$
$\Rightarrow 1 = A(t + 1)(3 + 2t) + B(t - 1)(3 + 2t) + C (t^2- 1)$
Put $t = -1$
$\Rightarrow 1 = -2B$ $\Rightarrow\text{A}=\frac{1}{10}$
Put $\text{t}=-\frac{3}{2}$
$\Rightarrow1=\frac{5}{4}\text{C}\Rightarrow\text{C}=\frac{4}{5}$
Thus,
$\text{I}=\frac{1}{10}\int\frac{\text{dt}}{\text{t}-1}-\frac{1}{2}\int\frac{\text{dt}}{\text{t}+1}+\frac{5}{4}\int\frac{\text{dt}}{3+2\text{t}}$
$=\frac{1}{10}\log|\text{t}-1|=\frac{1}{2}\log|\text{t}+1|+\frac{2}{5}\log|3+2\text{t}|+\text{C}$
Hence,
$\text{I}=\frac{1}{10}\log|\cos\text{x}-1|-\frac{1}{2}\log|\cos\text{x}+1|+\frac{2}{5}\log|3+2\cos\text{x}|+\text{C}$
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