What is #f(x) = int xsinx- sec2x dx# if #f(pi/12)=-2 #?

Question

What is #f(x) = int xsinx- sec2x   dx# if #f(pi/12)=-2 #?

Cameron Alexander · Accepted Answer

#color(blue)(f(x)=-x*cos x+sin x-1/2 ln (sec 2x+tan 2x) + 1/2 ln sqrt3+pi/24*sqrt(2+sqrt3)-1/2*sqrt(2-sqrt3)-2)#

also #color (blue)(f(x)=-x*cos x+sin x-1/2 ln (sec 2x+tan 2x) -1.73129)# Start by performing #color(blue)(" Integration by parts")# using the formula #color(blue)(int u dv=uv-int v du )# the solution is quite long because it consists of two terms. from the given: #int (x sin x - sec 2x) dx# #color (red)("First part:")# #int (x sinx) dx# Let #u=x#, and #dv=sin x dx# and #v=-cos x# and #du=dx# so that #int (x sinx) dx=-x*cos x-int -cos x dx# then #color (blue)(int (x sinx) dx=-x*cos x+sin x)# #color (red)(" second part:")# Use the formula: #int (sec u) du=ln (sec u+tan u)+C# so that #color(blue)(int sec 2x dx=1/2*ln(sec 2x+tan 2x))# Now , combine the first and second parts #color (magenta)(f(x)=int (x sin x - sec 2x) dx=# #color(magenta)(-x*cos x+sin x-1/2*ln(sec 2x+tan 2x)+C)# Solve for C: using #f(pi/12)=-2# and then #-2=-(pi/12)*cos (pi/12)+sin (pi/12)-1/2*lnabs(sec 2*(pi/12)+tan 2*(pi/12))+C# #-2=-(pi/12)*1/2*sqrt(2+sqrt3)+1/2*sqrt(2-sqrt3)-1/2*lnabs(2/sqrt3+1/sqrt3)+C# #C=1/2 ln sqrt3+pi/24*sqrt(2+sqrt3)-1/2*sqrt(2-sqrt3)-2# The final answer is #color (magenta)(f(x)=-x*cos x+sin x-1/2 ln (sec 2x+tan 2x) + 1/2 ln sqrt3+pi/24*sqrt(2+sqrt3)-1/2*sqrt(2-sqrt3)-2)# simplifying the square roots we have the following #color (magenta)(f(x)=-x*cos x+sin x-1/2 ln (sec 2x+tan 2x) -1.73129)#

Aurora Alvarado · Answer

undefined To find $ f(x) $, integrate $ \int xsin(x) - \sec^2(x) \, dx $. Given $ f(\frac{\pi}{12}) = -2 $, you can solve for $ f(x) $ using the given value and integration.