Question

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

Answer 1

`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`