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प्रश्न
`int (sinx)/(1 + sin x) "d"x`
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उत्तर
Let I = `int (sinx)/(1 + sin x) "d"x`
= `int (sinx)/(1 + sinx)* (1 - sinx)/(1 - sinx) "d"x`
= `int (sinx - sin^2x)/(1 - sin^2x) "d"x`
= `int (sinx - sin^2x)/(cos^2x) "d"x`
= `int((sinx)/(cos^2x) - (sin^2x)/(cos^2x)) "d"x`
= `int(1/(cosx) * (sinx)/(cosx) - tan^2x) "d"x`
= `int[secx tanx - (sec^2x - 1)] "d"x`
= `int secx tanx "d"x - int sec^2x "d"x + int1* "d"x`
∴ I = sec x − tan x + x + c
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