Question Statement
Evaluate the integral:
∫(1+tan2θ)dθ
Background and Explanation
This problem requires familiarity with fundamental trigonometric identities relating tangent and secant functions, as well as basic integration rules for trigonometric expressions.
Solution
We begin by splitting the integral into two separate terms using the linearity property:
I=∫(1+tan2θ)dθ=∫1dθ+∫tan2θdθ
The first integral evaluates to θ. For the second integral, we apply the trigonometric identity tan2θ=sec2θ−1:
I=θ+∫(sec2θ−1)dθ
Distributing the integral across the subtraction:
I=θ+∫sec2θdθ−∫1dθ
Evaluating each integral separately:
- ∫sec2θdθ=tanθ (standard integral)
- ∫1dθ=θ
Substituting these results back:
I=θ+tanθ−θ+c
Simplifying by combining like terms (θ−θ=0):
I=tanθ+c
Alternatively, we can simplify the integrand immediately using the identity:
sec2θ−tan2θ=1⇒1+tan2θ=sec2θ
Therefore:
I=∫(1+tan2θ)dθ=∫sec2θdθ=tanθ+c
To confirm our answer is correct, we differentiate the result:
dθd(tanθ+c)=dθd(tanθ)+dθd(c)=sec2θ+0=1+tan2θ
Since the derivative equals the original integrand, the solution tanθ+c is verified.
- Pythagorean Identity: 1+tan2θ=sec2θ (or equivalently sec2θ−tan2θ=1)
- Basic Integral: ∫sec2θdθ=tanθ+c
- Basic Integral: ∫1dθ=θ+c
- Linearity of Integration: ∫[f(θ)+g(θ)]dθ=∫f(θ)dθ+∫g(θ)dθ
Summary of Steps
- Identify the integrand: Recognize 1+tan2θ as a candidate for trigonometric simplification
- Apply identity: Use 1+tan2θ=sec2θ to rewrite the integrand (or split the integral and substitute tan2θ=sec2θ−1)
- Integrate: Evaluate ∫sec2θdθ=tanθ
- Add constant: Include the constant of integration +c
- Verify (optional): Differentiate the result to confirm it matches the original integrand