What is #f(x) = int xsin2x-6cotx dx# if #f(pi/4)=1 #?

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Answer 1 · 2018-02-09T19:06:55

#f(x)=-x/2*cos2x+1/4*sin2x-6ln(sinx)+1+(pisqrt2)/16-3Ln2#

Due to #int xsin2x*dx=x*(-1/2*cos2x)-int (-1/2)*cos2x*dx#

=#-x/2*cos2x+1/2intcos2x*dx#

=#-x/2*cos2x+1/4sin2x+C#

#f(x)=int (xsin2x-6cotx)*dx=-x/2*cos2x+1/4*sin2x-6ln(sinx)+C#

After imposing #f(pi/4)=1# condition,

#-pi/8*sin(pi/4)-6Ln(sin(pi/4))+C=1#

#-(pisqrt2)/16-6Ln(sqrt2/2)+C=1#

#-(pisqrt2)/16+3Ln2+C=1#

#C=1+(pisqrt2)/16-3Ln2#

Thus,

#f(x)=-x/2*cos2x+1/4*sin2x-6ln(sinx)+1+(pisqrt2)/16-3Ln2#