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प्रश्न
`int (sin(x - "a"))/(cos (x + "b")) "d"x`
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उत्तर
Let I = `int (sin(x - "a"))/(cos (x + "b")) "d"x`
= `int (sin[(x + "b") - ("a" + "b")])/(cos(x + b) "d"x`
= `int (sin(x + "b")*cos("a" + "b") - cos(x + "b")*sin("a" + "b"))/(cos(x + "b")) "d"x`
= `int[(sin(x + "b")*cos("a" + "b"))/(cos(x + "b")) - (cos(x + "b")*sin("a" + "b"))/(cos(x + "b"))] "d"x`
= `int [tan (x + "b")*cos("a" + "b") - sin("a" + "b")] "d"x`
= `cos("a" + "b") int tan(x + "b")* "d"x - sin("a" + "b") int "d"x`
∴ I = cos (a + b). log |sec (x + b)| – [sin (a + b)] x + c
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