∫cosx1t2f(t)dt=sin3x+cosx
On differentiating, we get
f(cosx)sinx⋅cos2x=3sin2xcosx−sinx
f(cosx)=3tanx−sec2x
Again differentiating, we get
−sinxf′(cosx)=3sec2x−2sec2xtanx
When cosx=31 then secx=3, \mathrm{tan}x=\sqrt{2}&\mathrm{sin}x=\frac{\sqrt{2}}{\sqrt{3}}
Then −32f′(31)=3×3−2×32
⇒31f′(31)=6−29