`int (sin(x - "a"))/(cos (x + "b")) "d"x`

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` &there4; I = cos (a + b). log |sec (x + b)| – [sin (a + b)] x + c

∫sin(x-a)cos(x+b) dx - Mathematics and Statistics

प्रश्न

$\int \frac{\sin (x - a)}{\cos (x + b)} d x$

योग

उत्तर

Let I = $\int \frac{\sin (x - a)}{\cos (x + b)} d x$

= $\int \frac{\sin [(x + b) - (a + b)]}{\cos (x + b) d x}$

= $\int \frac{\sin (x + b) \cdot \cos (a + b) - \cos (x + b) \cdot \sin (a + b)}{\cos (x + b)} d x$

= $\int [\frac{\sin (x + b) \cdot \cos (a + b)}{\cos (x + b)} - \frac{\cos (x + b) \cdot \sin (a + b)}{\cos (x + b)}] d x$

= $\int [\tan (x + b) \cdot \cos (a + b) - \sin (a + b)] d x$

= $\cos (a + b) \int \tan (x + b) \cdot d x - \sin (a + b) \int d x$

∴ I = cos (a + b). log |sec (x + b)| – [sin (a + b)] x + c

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Methods of Integration: Integration by Parts

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अध्याय 2.3: Indefinite Integration - Short Answers I

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अध्याय 2.3 Indefinite Integration
Short Answers I | Q 5

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Methods of Integration: Integration by Parts

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