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Question
Find `int_ sin ("x" - a)/(sin ("x" + a )) d"x"`
Solution
Let I = `int_ sin("x" - a)/sin ("x" + a)d"x"`
⇒ I = `int_ sin [("x" + a) - 2a]/sin ("x" + a)d"x"`
= `int_ (sin ("x" + a )·cos (2a) - cos ("x" + a)· sin (2a))/sin ("x" + a)d"x"`
= `int_ cos (2a) d"x" - int_ cot ("x" + a)· sin (2a)d"x"`
= x·cos (2a) - log|sin (x + a)|·sin (2a) + C
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