Question
Evalute the following integrals:
$\int\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\text{dx}$
$\int\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\text{dx}$
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Answer
Let $\text{I}=\int\frac{\sin(\text{x}-\text{a})}{\sin(\text{x}-\text{b})}\text{dx}$
Putting x - b = t
⇒ x = b + t
& dx = dt
$\therefore\text{I}=\int\frac{\sin(\text{b}+\text{t}-\text{a})}{\sin\text{t}}\text{dt}$
$=\int\frac{\sin\big\{(\text{b}-\text{a})+\text{t}\big\}}{\sin\text{t}}\text{dt}$
$=\int\frac{\sin(\text{b}-\text{a})\cos\text{t}}{\sin\text{t}}+\int\frac{\cos(\text{b}-\text{a})\sin\text{t}}{\sin\text{t}}\text{dt}$
$=\int\sin(\text{b}-\text{a})\cot\text{t dt}+\int\cos(\text{b}-\text{a})\text{dt}$
$=\sin(\text{b}-\text{a})\text{in}|\sin\text{t}\big|+\text{t}\cos(\text{b}-\text{a})+\text{C}$
$=\sin(\text{b}-\text{a})\text{in}|\sin(\text{x}-\text{b})|+(\text{x}-\text{b})\cos(\text{b}-\text{a})+\text{C}$
$\big[\because\text{t}=\text{x}-\text{b}\big]$
Putting x - b = t
⇒ x = b + t
& dx = dt
$\therefore\text{I}=\int\frac{\sin(\text{b}+\text{t}-\text{a})}{\sin\text{t}}\text{dt}$
$=\int\frac{\sin\big\{(\text{b}-\text{a})+\text{t}\big\}}{\sin\text{t}}\text{dt}$
$=\int\frac{\sin(\text{b}-\text{a})\cos\text{t}}{\sin\text{t}}+\int\frac{\cos(\text{b}-\text{a})\sin\text{t}}{\sin\text{t}}\text{dt}$
$=\int\sin(\text{b}-\text{a})\cot\text{t dt}+\int\cos(\text{b}-\text{a})\text{dt}$
$=\sin(\text{b}-\text{a})\text{in}|\sin\text{t}\big|+\text{t}\cos(\text{b}-\text{a})+\text{C}$
$=\sin(\text{b}-\text{a})\text{in}|\sin(\text{x}-\text{b})|+(\text{x}-\text{b})\cos(\text{b}-\text{a})+\text{C}$
$\big[\because\text{t}=\text{x}-\text{b}\big]$
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