Question

Evaluate:

`int (sin"x"+cos"x")/(sqrt(9+16sin2"x")) "dx"`

Answer 1

`"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"`