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Use integration by parts to prove the reductionformula.$$\begin{array}{l}{\int \sec ^{n} x d x=\frac{\tan x \sec ^{n-2} x}{n-1}+\frac{n-2}{n-1} \int \sec ^{n-2} x d x} \\ {(n \neq 1)}\end{array}$$

$\frac{1}{n-1} \sec ^{n-2} x \tan x+\frac{n-2}{n-1} \int \sec ^{n-2} x d x$

Calculus 2 / BC

Chapter 6

TECHNIQUES OF INTEGRATION

Section 1

Integration by Parts

Integration Techniques

Improper Integrals

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Section six That one problem 38 here were asked ever derived the reduction formula for how to integrate the seeking to the power and said to give us the formula that were asked to arrive here. So let's start out with looking at the in a row of seeking to the power of X DX. Let's just write this as the integral of seeking squared of X times. A seeking of in mine It's too of X dx. Now let's do integration by parts. So let's let you equal the seeking to the n minus two Power of X. Then d'you is going to be in minus two times this sequent to the n minus three of x times, the derivative of the seeking of eggs which is just seeking Innovex tangent of X. So all of that just using the chain role. So I've got you Do you Devi is going to be seeking squared of x dx. Therefore, V is going to be the tangent of X So use of this Teoh integrate by parts So I come up with the integral of the Sikh into the power of X dx is going to be u v So that's going to be change it of X seeking to the n minus two of acts that you ve minus the integral of V D u so minus the in a row. So you're gonna have in minus two and then you're going to have seek it to the N minus three of X and then you're going to have seeking X change in X tangent X dx. Okay, so we'll clean all of this up. So I've got the integral of the Sikh into the inthe power of X DX is equal to the tangent of X seeking to the N minus two of X minus in minus two. And then it looks like I've got, um so seeking to the n minus three temps of seeking of X, that's just going to be seeking to the n minus two of X and then there to tangents there. So changes squared of X DX. Now I can use my Pythagorean identity. I know that the tangent squared isjust what that is the seeking squared of X minus one. So this becomes the in a row of seeking to the power of X DX is equal to tangent of X seeking to the n minus two of X minus in minus two. And then the integral just become seeking to the n minus two of X times seeking squared of X minus one DX. Now I can break that integral into two separate intervals. So this becomes the interval of the seeking to the power of X DX is equal to Tangent of X seeking to the n minus two of acts minus in minus two. So when you multiply the seeking to the n minus two in this sequence square, that's just going to give you seeking to the 10th power So seek into the end power of X and then minus in minus two. And then when you multiply the seeking to the n minus two and a minus one, you're gonna get plus seeking to the n minus two of X DX. Now, what you notice is this integral. It's the same on both sides of the equation, so I can move that to the left side of equation. So you're gonna have the integral seeking to the power of X DX plus in minus two. Integral seeking in power of X DX is equal to change in X, seek it in minus two of X and then plus in minus two seeking to the n minus two of X dx. Now what happens when you had one plus in minus two? That gives you in minus one. The integral seeking to the power of X dx is equal to tangent of X seeking to the n minus two of acts plus in minus two seeking to the n minus two of X t x And now in order to solve for this interval, you divide everything by in minus one. So the integral seeking to the power of X dx is going to be equal to Tangent of X, seeking to the n minus two of X, divided by n minus one plus in minus two over in minus one Sequent to the n minus two of X dx and that's our final value here. So again, the key here was many times and in arose is to look at the integral, different way. So look at seeking to the end as seeking squared seeking in minutes to the next key was to do integration by parts integration by parts allowed me to solve for this interval and come up with the solution for the reduction formula

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