What is #f(x) = int xsin2x+cosx dx# if #f(pi/2)=-4 #?

1 Answer
sjc
Apr 24, 2018

see below

Explanation:

#I=int(xsin2x+cosx)dx#

we have two integrals, the first one needs to be done by parts, the second is a standard integral. We will omit the constant of integration until the end.

#f(x)=I=int(xsin2x)dx+intcosxdx=I_1+I_2#

#I_2=intcosxdx=sinx--(1)#

#I_1=intxsin2xdx#

#IBP#

#intu(dv)/(dx)dx=uv-intv(du)/(dx)dx#

#u=x=>(du)/(dx)=1#

#(dv)/(dx)=sin2x=>v=-1/2cos2x#

#I_1=-1/2xcos2x+int1/2cos2xdx#

#I_1=-1/2xcos2x+1/4sin2x#

#:.I=I-1+I_2=f(x)=-1/2xcos2x+1/4sin2x+sinx+c#

now#f(pi/2)=-4#

#-4=-1/2pi/2cos2(pi/2)+1/4sin(2(pi/2))+sin(pi/2)+c#

#-4=-(pi/4)cancel(cos(pi))^(-1)+cancel(1/4sinpi)^0+cancel(sin(pi/2))^1+c#

#=>-4=pi/4+1+c#

#c=-5-pi/4#

#f(x)=-1/2xcos2x+1/4sin2x+sinx-5-pi/4#

Related questions
See all questions in Evaluating the Constant of Integration
Impact of this question
1310 views around the world
You can reuse this answer
Creative Commons License