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

Question

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

Paisley Johnson · Accepted Answer

#f(x)=xsinx+cosx-2-1/sqrt2(pi/4+1)#.

#f(x)=intxcosxdx# We have to integrate by using, the Rule of Integration by Parts : #intuvdx=uintvdx-int((du)/dxintvdx)dx# We take #u=x rArr (du)/dx=1#, and, #v=cosx rArr intvdx=sinx# Hence, #f(x)=xsinx-intsinxdx=xsinx+cosx+C# To determine the const. of Integration #C#, we use, #f(pi/4)=-2 rArr pi/4sin(pi/4)+cos(pi/4)+C=-2# # rArr pi/4*1/sqrt2+1/sqrt2+C=-2# #rArr C=-2-1/sqrt2(pi/4+1)# Therefore, #f(x)=xsinx+cosx-2-1/sqrt2(pi/4+1)#.

Isaiah Allen · Answer

undefined To find $ f(x) = \int x \cos(x) \, dx $ when $ f(\frac{\pi}{4}) = -2 $, we can first differentiate $ f(x) $ using integration by parts:

$$ \int u \, dv = uv - \int v \, du $$

Let $ u = x $ and $ dv = \cos(x) \, dx $, then $ du = dx $ and $ v = \sin(x) $.

$$ \int x \cos(x) \, dx = x\sin(x) - \int \sin(x) \, dx $$

$$ = x\sin(x) + \cos(x) + C $$

Given that $ f(\frac{\pi}{4}) = -2 $, we can plug this into the equation:

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

Solve for $ C $:

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

$$ = -2 - \frac{\pi}{4}\left(\frac{\sqrt{2}}{2}\right) - \frac{\sqrt{2}}{2} $$

$$ = -2 - \frac{\pi\sqrt{2}}{8} - \frac{\sqrt{2}}{2} $$

$$ = -2 - \frac{\pi\sqrt{2} + 4\sqrt{2}}{8} $$

$$ = -2 - \frac{\sqrt{2}(\pi + 4)}{8} $$

Therefore, $ f(x) = x\sin(x) + \cos(x) - 2 - \frac{\sqrt{2}(\pi + 4)}{8} $.