# What is #f(x) = int xsinx dx# if #f(pi/4)=-2 #?

just IBP its is easiest way

Here

So we have

now using the IV

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To find ( f(x) = \int x \sin(x) , dx ) if ( f\left(\frac{\pi}{4}\right) = -2 ), we'll integrate ( x \sin(x) ) and use the given condition to determine the constant of integration. Integrating ( x \sin(x) ) by parts:

[ u = x, \quad dv = \sin(x) , dx ] [ du = dx, \quad v = -\cos(x) ]

[ \int x \sin(x) , dx = -x \cos(x) - \int -\cos(x) , dx ] [ = -x \cos(x) + \sin(x) + C ]

Given ( f\left(\frac{\pi}{4}\right) = -2 ), substitute ( \frac{\pi}{4} ) into ( f(x) ) and solve for ( C ):

[ -2 = -\frac{\pi}{4} \cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right) + C ]

[ -2 = -\frac{\pi}{4} \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} + C ] [ -2 = -\frac{\pi}{4\sqrt{2}} + \frac{\sqrt{2}}{2} + C ] [ -2 = -\frac{\pi}{4\sqrt{2}} + \frac{\sqrt{2}}{2} + C ]

[ C = -2 + \frac{\pi}{4\sqrt{2}} - \frac{\sqrt{2}}{2} ]

So, ( f(x) = -x \cos(x) + \sin(x) - 2 + \frac{\pi}{4\sqrt{2}} - \frac{\sqrt{2}}{2} ).

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