Question

Verify the following reduction formula:
$\displaystyle \int \sec^n(u)\, du=\frac{\sec^{n-2}(u)\tan(u)}{n-1}+\frac{n-2}{n-1}\int \sec^{n-2}(u)\, du, \; n\neq 1$

Answer 1

Answer:

See Explanation.

General Formulas and Concepts:

Pre-Algebra

• Distributive Property
• Equality Properties

Algebra I

• Combining Like Terms

Algebra II

• Exponential Rules

Pre-Calculus

• Pythagorean Identities: tan²(x) = sec²(x) - 1

Calculus

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Integration Rule 1:         $\int {cf(x)} \, dx = c\int {f(x)} \, dx$

Integration Rule 2:        $\int {f(x) \pm g(x)} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Integration 1:                 $\int {sec^2(u)} \, du = tan(u) + C$

Integration by Parts:    $\int {u} \, dv = uv - \int {v} \, du$

• [IBP] LIPET: Logs, inverses, Polynomials, Exponentials, Trig

Step-by-step explanation:

Step 1: Define

$\int {sec^n(u)} \, du$

Step 2: Rewrite

1. [Integral - Alg] Separate Exponents:                    $\int {sec^n(u)} \, du = \int {sec^{n-2}(u)sec^2(u)} \, du$

Step 3: Identify Variables

Using LIPET, we define variables to use IBP.

Use Integration 1.

$u = [sec(u)]^{n-2} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dv = sec^2(u)du\\du = (n-2)[sec(u)]^{n-3} sec(u)tan(u) \ \ \ \ \ \ \ \ v = tan(u)$

Step 4: Integrate

1. Integrate [IBP]:   $\int {sec^n(u)} \, du = tan(u)[sec(u)]^{n-2} - \int [{tan(u)(n-2)[sec(u)]^{n-3}sec(u)tan(u)} ]\, du$
2. [Integral - Alg] Multiply:                  $\int {sec^n(u)} \, du = tan(u)[sec(u)]^{n-2} - \int [{tan^2(u)(n-2)[sec(u)]^{n-2}}] \, du$
3. [integral - Int Rule 1] Simplify:     $\int {sec^n(u)} \, du = tan(u)[sec(u)]^{n-2} - (n-2)\int [{tan^2(u)[sec(u)]^{n-2}}] \, du$
4. [Integral - Pythagorean Identities] Rewrite:   $\int {sec^n(u)} \, du = tan(u)[sec(u)]^{n-2} - (n-2)\int [{[sec^2(u) - 1][sec(u)]^{n-2}}] \, du$
5. [Integral - Alg] Multiply/Distribute:   $\int {sec^n(u)} \, du = tan(u)[sec(u)]^{n-2} - (n-2)\int [{sec^n(u)-[sec(u)]^{n-2}}] \, du$
6. [Integral - Int Rule 2] Rewrite:     $\int {sec^n(u)} \, du = tan(u)[sec(u)]^{n-2} - (n-2) [\int {sec^n(u)} \, du - \int {[sec(u)]^{n-2}} \, du ]$
7. [Integral - Alg] Distribute:   $\int {sec^n(u)} \, du = tan(u)[sec(u)]^{n-2} - (n-2) \int {sec^n(u)} \, du + (n-2)\int {[sec(u)]^{n-2}} \, du$
8. Rewrite:     $\int {sec^n(u)} \, du = sec^{n-2}(u)tan(u) - (n-2) \int {sec^n(u)} \, du + (n-2)\int {[sec(u)]^{n-2}} \, du$
9. [Integral - Alg] Isolate Integral Term:     $\int {sec^n(u)} \, du + (n-2) \int {sec^n(u)} \, du = sec^{n-2}(u)tan(u) + (n-2)\int {[sec(u)]^{n-2}} \, du$
10. [Integral - Alg] Combine Like Terms:     $(n - 1)\int {sec^n(u)} \, du = sec^{n-2}(u)tan(u) + (n-2)\int {[sec(u)]^{n-2}} \, du$
11. [Integral 2 - Alg] Rewrite:     $(n - 1)\int {sec^n(u)} \, du = sec^{n-2}(u)tan(u) + (n-2)\int {sec^{n-2}(u)} \, du$
12. [Integral - Alg] Isolate Original Integral:     $\int {sec^n(u)} \, du = \frac{sec^{n-2}(u)tan(u)}{n-1} + \frac{n-2}{n-1} \int {sec^{n-2}(u)} \, du$

And we have proved the Reduction Formula!