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

Question

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

Evelyn Agard · Accepted Answer

#= - x cos x + sin x + 1/sqrt 2 (pi/4 -1) - 2#

#f(x) = int dx \ (x \ sinx)#

just IBP its is easiest way

Here
#u = x , u' =1 #

#v' = sin x, v = - cos x#

So we have

#- x cos x - int dx \ (- cos x * 1)#

#= - x cos x + int dx \ ( cos x)#

#= - x cos x + sin x + C#

now using the IV

#-2 = - pi/4 1/sqrt 2 + 1/sqrt 2 + C#

# C = 1/sqrt 2 (pi/4 -1) -2 #

Luke Alvarez · Answer

undefined 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} $.