Question

the curve y = sin^2 2x cos x for 0 ≤ x ≤1/2π and its maximum point M.

(i) Find the x-coordinate of M.

(ii) Using the substitution u = sin x, find by integration the area of the shaded region bounded by the curve and the x-axis.

Answer 1

find max by setting dy/dx = 0
you will have to use product and chain rules to get derivative

Answer 2

\[y' = (\sin^2 2x)' \cos x + \sin^2 2x (\cos x)'=0\]
\[y' = 4 \sin 2x \cos 2x \cos x - \sin^2 2x \sin x = 0\]

Answer 3

did that help?
can you solve it from here?

Answer 4

\(\large\tt \color{black}{4sin2xcos2xcosx-sin^22xsinx=0}\)

\(\large\tt \color{black}{4cos2xcosx-sin2xsinx=0}\)

\(\large\tt \color{black}{4(cos^2x-sin^2x)cosx-(2sinxcosx)sinx=0}\)

\(\large\tt \color{black}{4(cos^3x-sin^2xcosx)-(2sin^2xcosx)=0}\)

\(\large\tt \color{black}{4(cos^3x-sin^2xcosx)-(2sin^2xcosx)=0}\)

\(\large\tt \color{black}{4(cos^2x-sin^2x)-(2sin^2x)=0}\)

\(\large\tt \color{black}{4cos^2x-4sin^2x-2sin^2x=0}\)

\(\large\tt \color{black}{4cos^2x-6sin^2x=0}\)

\(\large\tt \color{black}{6sin^2x=4cos^2x}\)

\(\large\tt \color{black}{\dfrac{sin^2x}{cos^2x}=\dfrac{4}{6}}\)

\(\large\tt \color{black}{tan^2x=\dfrac{4}{6}=\dfrac{2}{3}}\)

\(\large\tt \color{black}{tanx=\pm\sqrt{\dfrac{2}{3}}}\)

we will consider only \(\large\tt \color{black}{ tanx=+\sqrt{\dfrac{2}{3}}}\) as 0 ≤ x ≤1/2π


\(\large\tt \color{black}{x=\arctan{\sqrt{\dfrac{2}{3}}}}\)

\(\large\tt \color{black}{x=0.684}\)