#### To determine

**To evaluate:**

The given indefinite integral ∫tan2θsec2θdθ

#### Answer

∫tan2θsec2θdθ= 13tan3θ+C

From the graph, it is reasonable to say that 13tan3θ+C is antiderivative of tan2θsec2θ

#### Explanation

**1) Concept:**

i) The substitution rule:

If u=g(x) is a differentiable function whose range is I and f is continuous on I, then ∫f(gx)g'xdx=∫f(u)du. ii) Indefinite integral

∫xn dx=xn+1n+1+C n≠-1

**2) Given:**

∫tan2θsec2θdθ

**3) Calculation:**

The given integral is ∫tan2θsec2θdθ. Here, use the substitution method.

Substitute tanθ=u.

Differentiation with respect to θ.

sec2θdθ=du

Therefore, the given integral becomes

∫tan2θsec2θdθ=∫u2du

Using concept ii),

=u2+12+1+C

=u33+C

=13u3+C

Resubstitute tanθ=u.

Hence the solution become

∫tan2θsec2θdθ=13tan3θ+C

Now, take C=0 and sketch the graph of the function fθ=tan2θsec2θ and its antiderivative Fθ=13tan3θ

The red graph denotes the function fθ=tan2θsec2θ and the blue graph denotes the function Fθ=13tan3θ

The answer is reasonable since, f(θ) is always positive and Fθ is always increasing. Also at zeros of f(θ), Fθ has a horizontal tangent.

**Conclusion:**

Therefore,

∫tan2θsec2θdθ= 13tan3θ+C

From the graph, it is reasonable to say 13tan3θ+C is antiderivative of tan2θsec2θ