To determine

To evaluate:

∫π/6πsinθ dθ

Answer

∫π/6πsinθ dθ=1+32

Explanation

1) Concept:

The Fundamental Theorem of Calculus(part 2) : Suppose f is continuous on [a, b], then

∫abfxdx=Fb-F(a), where F is antiderivative of  f, that is F'=f.

2) Calculation:

By using the antiderivative of sin⁡θ and fundamental theorem of calculus

∫π/6πsinθ dθ=-cosθππ/6

=-cos⁡π--cos⁡π6=-cos⁡π+cos⁡π6

After substituting values of cos⁡(π) and cos⁡π6,

∫π/6πsinθ dθ=--1+32=1+32

Conclusion:

Therefore,

∫π/6πsinθ dθ=1+32