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प्रश्न
Evaluate:
`int (sin"x"+cos"x")/(sqrt(9+16sin2"x")) "dx"`
उत्तर
`"I" =int (sin"x"+cos"x")/(sqrt(9+16sin2"x")) "dx"`
Let sin x - cos x = t
(cos x + sin x) dx = dt
(sin x - cos x)2 = t2
sin2x + cos2x - 2 sin x cos x = t2
⇒ 1 - sin 2x = t2
sin 2x = 1 - t2
`"I" = int "dt"/sqrt(9+16(1-"t"^2)`
` = int "dt"/sqrt(9+16-16t^2)`
` = int "dt"/sqrt(25-16t^2)`
`= int "dt"/sqrt(16(25/16-"t"^2))`
`=1/4 int "dt"/(sqrt((5/4)^2)-t^2)`
`= 1/4sin^-1 "t"/((5/4))`
`=1/4 sin^-1 ((4"t")/5)`
`"I" = 1/4 sin^-1 ((4(sin"x"-cos"x"))/2) + "C"`
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