74. A secant reduction formula
Prove that for positive i...

Question

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.)

Accepted Answer

Welcome back, everyone. Find the reduction formula for the integral of tangent to the power of N of XDX for positive integers and not equal to 1. For this problem, since we're going to identify the reduction formula, we can begin by analyzing the integra. We have tangent to the power of N of X. We can split tangent squared to begin with, and we can do that by using the properties of exponents. We can simply write tangent. Squared of x multiplied by tangent to the power of M minus 2. Of X, right? Because according to the properties of exponents, when we multiply the same base with different exponents, we add them and minus 2 + 2 gives us the original N value. So now we can also recall that. E squared of X can be written as C2 x minus 1. This is going to be useful for substitution. We can now show that. The original integral tangents. To the power of N of XD X. Can be written as the integral. Of tangent to the power of n minus 2 of X multiplied by. Second squared off x minus 1 d X. Now, let's apply the distributive property to get two integrals. One of them is going to be tangent trace to the power of N minus 2 of X. C squad of XD X and then we're subtracting the integral of tangent to the power of n minus 2 of XD X. We can focus on the first integral and we can use the U substitution. Let's suppose that U equals tangent of X, which essentially means that D U is equal to C2XDX using the derivative of tangent, right? If we continue our equation, the first integral can be written as you raise the power of N minus 2 because now tangent of X becomes U. Incred X T X becomes Tu. For the second integral, we can just leave it as it is. So we're subtracting the integral of tangent to the power of N minus 2 of XD X. And this is because we're looking for a reduction formula, so we're going to leave that integral. And now we can integrate the first part. U to the power of N minus 2D you can be integrated using the power rules, so we're going to use U. Erase the power of n minus 2 + 1 gives us N minus 1 and we're dividing by that new exponent N minus 1. And subtracting the integral of tangent to the power of n minus 2 of XD X. We don't need to include any exponent. I'm sorry, we don't need to include any constants of integration because we still have one integral remaining, right? And now, finally, let's replace you with tangent of X. So we get tangent. To the power of n minus 1 of x divided by n minus 1 minus the integral of tangent to the power of n minus 2 of X D X where n is not equal to 1 and this is going to be our final reduction formula. Thank you for watching.

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.)

Key Concepts

Integration by Parts

Reduction Formulas

Trigonometric Identities and Derivatives

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