If #f' (u)- 4(cos u sin 2u)#, and #f(pi/6) = 11#, what is #f(u)#?

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If #f' (u)=4(cos u sin 2u)#, and #f(pi/6) = 11#, what is #f(u)#?

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Answer 1 · 2017-12-04T05:03:43

# f(u)=-2/3cos3u-2cosu+11+sqrt3.#

Since, #f'(u)=4sin2ucosu=2{2sin2ucosu},#

#:. f'(u)=2{sin(2u+u)+sin(2u-u)}, or,#

#f'(u)=2(sin3u+sinu).#

Recall that, #g'(x)=G(x) rArr g(x)=intG(x)dx+c.#

#:. f(u)=intf'(u)du,#

#=int2(sin3u+sinu)du,#

#=2{-(cos3u)/3-cosu}+C, or,#

#f(u)=-2/3cos3u-2cosu+C.#

But, given that,

#f(pi/6)=11rArr-2/3cos(3*pi/6)-2cos(pi/6)+C=11.#

#rArr -2/3cos(pi/2)-2(sqrt3/2)+C=11,#

#rArr 0-sqrt3+C=11,#

# rArr C=11+sqrt3.#

#"Therefore, "f(u)=-2/3cos3u-2cosu+11+sqrt3.#