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,

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

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