I=12∫01+√3((x+1)2(1−x)6)14 dx=12∫01+√3(1−x)√1−x2dx
Let x=sinθ⇒dx=cosθ dθ
I=π6∫0(1+√3)cosθcosθ(1−sinθ)dθ=(1+√3)π6∫0(1+sinθ)cos2θdθ
I=(1+√3)⎛⎜⎝π6∫0sec2θ dθ+π6∫0secθtanθ dθ⎞⎟⎠
I=(1+√3)(tanθ∣∣π60+secθ∣∣π60)
I=(1+√3)(1√3+(2√3−1))
I=(√3+1)(√3−1)=((√3)2−12)=2