What is #f(x) = int xsin2x+cos3x dx# if #f(pi/12)=4 #?

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Answer 1 · 2017-10-28T15:13:14

#f(x) = -x/2cos(2x)+1/4sin(2x)+1/3 sin(3x) + 1/48 (186 - 8 sqrt(2) + sqrt(3) pi)#

Given: #f(x) = int xsin(2x)+cos(3x) dx; f(pi/12) = 4#

#f(x) = int xsin2xdx+int cos3x dx#

Setup the integration by parts variable for the first integral:

Let #u = x and dv = sin(2x)dx#
Then #du = dx# and #v = -1/2cos(2x)dx#

#f(x) = -x/2cos(2x)+1/2int cos(2x)dx+int cos3x dx#

The remaining integrals become sine functions multiplied by a constant:

#f(x) = -x/2cos(2x)+1/4sin(2x)+1/3 sin(3x) + C#

To find the value of C, evaluate at the point #(pi/12,4)#

#4 = -(pi/12)/2cos(2(pi/12))+1/4sin(2(pi/12))+1/3 sin(3(pi/12)) + C#

I let WolframAlpha sort this one out:

#C = 1/48 (186 - 8 sqrt(2) + sqrt(3) pi)#

Substitute the value for C into #f(x)#:

#f(x) = -x/2cos(2x)+1/4sin(2x)+1/3 sin(3x) + 1/48 (186 - 8 sqrt(2) + sqrt(3) pi)#