What is f(x) = int sinx-cos^2x dx if f(pi/6)=2 ?

1 Answer
Mar 10, 2018

f(x)=-[cosx+x/2+1/4*sin2x]+(48+15sqrt(3)+2pi)/24

Explanation:

f(x)=int(sinx-cos^2x)dx
f(x)=int(sinx-1/2(1+cos2x))dx=[-cosx-1/2(x+(sin2x)/2)]+c
f(x)=[-cosx-x/2-1/4*sin2x]+c, to (1)
f(pi/6)=[-cos(pi/6)-pi/12-1/4*sin(pi/3)]+c=2=>[-sqrt(3)/2-pi/12-1/4sqrt(3)/2]+c=2=>-[sqrt(3)/2+pi/12+1/4sqrt(3)/2]+c=2=>-(12sqrt(3)+2pi+3sqrt(3))/24+c=2=>c=2+(15sqrt(3)+2pi)/24
.c=(48+15sqrt(3)+2pi)/24, to (2)
From (1) and (2)
f(x)=-[cosx+x/2+1/4*sin2x]+(48+15sqrt(3)+2pi)/24