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प्रश्न
Find: `int (sin2x)/sqrt(9 - cos^4x) dx`
बेरीज
उत्तर
Put `cos^2x` = t ⇒ `−2cosxsinxdx` = dt ⇒ `sin2xdx = -dt`
The given integral = `- int (dt)/sqrt(3^2 - t^2) = - sin^(-1) t/3 + c = - sin^(-1) (cos^2x)/3 + c`
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