Answer to Question #263809 in Trigonometry for kazbek

Question

y=ctgx+sin²x³

Answer 1

$y=\ctg x+\sin^2x^3$

$(\ctg x)'= -\dfrac1{\sin²\ x}$

$(\sin^2x)' = sin(2x)\\ \therefore a = (\sin^2x^3)' =(\sin x³ × \sin x³)'\\ u = \sin x³ ;\quad du = 3x²\cos x³\\ v = \sin x³ ;\quad dv= 3x²\cos x³\\ da/dx = vdu + udv\\ \therefore(\sin x³ × \sin x³)' = \\ 3x²\sin x³\cos x³ + 3x²\sin x³\cos x³ = 6x²\sin x³\cos x³$

$dy/dx = -\dfrac1{\sin^2 x}+ 6x² \sin x³\cos x³$