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(a) d(ln(sec³(2x)))/dx

= (1/sec³(2x)) ( 3 sec²(2x)) d(sec(2x))/dx

 = (3/sec(2x)) (2 sec(2x) (tan 2x))

= 6 tan(2x)

(b) At point a, y = -3

-3 = -3tan(2x)

x= π/8

Area bounded by the curve y = -3tan(2x) from x=0 to x=a is 

= ∫0 π/8 -3tan(2x) dx

= -3/6 * ( ln(sec³(π/4)) – ln(sec³(0))

= -1/2*   ln  (2)3/2

= -3/4  ln (2)                       proved

The area of the curve is 3/4 ln(2) . The negative sign merely exists because the area is below y =0.

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