Question

74. A secant reduction formulaProve that for positive integers n ≠ 1,∫ secⁿ x dx = (secⁿ⁻² x tan x)/(n − 1) + (n − 2)/(n − 1) ∫ secⁿ⁻² x dx.(Hint: Integrate by parts with u = secⁿ⁻² x and dv = sec² x dx.)

Accepted Answer

Start with the integral $$ I_n = \int \sec^n x \, dx $$ where $$ n  is a $$positive integer and $$ n 
eq 1 . $$According to the hint, use integration by parts with $$ u = \sec^{n-2} x $$ and $$ dv = \sec^2 x \, dx . $$Compute $$ du $$ and $$ v $$: 
- Differentiate $$ u $$: $$ du = (n-2) \sec^{n-3} x \sec x 	an x \, dx = (n-2) \sec^{n-2} x 	an x \, dx .
- $$Integrate $$ dv $$: $$ v = 	an x $$ since $$ \frac{d}{dx} (	an x) = \sec^2 x . $$Apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Substitute the expressions:

$$ I_n = \sec^{n-2} x 	an x - \int 	an x \cdot (n-2) \sec^{n-2} x 	an x \, dx . $$Simplify the integral inside:

$$ I_n = \sec^{n-2} x 	an x - (n-2) \int \sec^{n-2} x 	an^2 x \, dx .

$$Recall the identity $$ 	an^2 x = \sec^2 x - 1  to $$rewrite the integral: Rewrite the integral using the identity:

$$ I_n = \sec^{n-2} x 	an x - (n-2) \int \sec^{n-2} x (\sec^2 x - 1) \, dx $$

which expands to

$$ I_n = \sec^{n-2} x 	an x - (n-2) \int \sec^n x \, dx + (n-2) \int \sec^{n-2} x \, dx .

$$Now, isolate $$ I_n  on $$one side to express it in terms of $$ \int \sec^{n-2} x \, dx .$$

74. A secant reduction formula
Prove that for positive integers n ≠ 1,
∫ secⁿ x dx = (secⁿ⁻² x tan x)/(n − 1) + (n − 2)/(n − 1) ∫ secⁿ⁻² x dx.
(Hint: Integrate by parts with u = secⁿ⁻² x and dv = sec² x dx.)

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Key Concepts

Integration by Parts

Reduction Formulas

Trigonometric Identities and Derivatives