#### To determine

**To explain:**

What is wrong inthe given equation.

#### Answer

The given expression for ∫π/ 3πsecθtanθdθ is incorrect, since ∫π/ 3πsecθtanθdθ does not exist as a difference of anti-derivative.

#### Explanation

**1) Concept:**

Use the fundamental theorem of calculus part 2 to explain the wrong in the given equation.

**2) Fundamental theorem of Calculus, Part **2

If f is continuous on a,b, then

∫abfxdx=Fb-F(a)

where F is any antiderivative of f, that is, a function F such that F'=f

**2) Given:**

∫π/ 3πsecθtanθdθ=secθ]ππ/ 3=-3

**3) Calculation:**

The graph of function fθ=secθtanθ, π/3≤θ≤π is given by,

By trigonometric formula,

secθ=1cosθ, tanθ=sinθcosθ

Therefore,

fθ=secθtanθ=1cosθ·sinθcosθ=sinθcos2θ

Therefore, the function fθ=secθtanθ is not continuous on π/3,π

So, the Fundamental theorem of Calculus, Part 2 cannot be applied. So we cannot express the given integral as

∫π/ 3πsecθtanθdθ=Fπ-Fπ/3 where F is antiderivative of f, that is, a function F such that F'(θ)=f(θ). Besides the definite integral doesn’t exists.

**Conclusion:**

∫π/ 3πsecθtanθdθ does not exist as a difference of anti-derivative, therefore the given equation is wrong.