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Solve the given problems by integration.For $n > 0,$ show that $\int \tan x \sec ^{n} x d x=1 \sec ^{n} x+C$
Calculus 1 / AB
Chapter 28
Methods of Integration
Section 5
Other Trigonometric Forms
Integrals
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six at 1 37 were asked to derive the formula for the tangent to the power, the integration of that. So when I see this, I look at the integral as an in minus two. So how could I take often? The case with integration is can I move things around in a way to manipulate to get something? So what if I were to write this as the integral of the changes squared of X times a change it of in minus two of x dx. So if I do this, then the in a rope stays the same. But then I have a path Agrium identity. I know that one. Plus, the tensions squared of X is equal to the seeking squared of X. So I could rewrite all of this as the integral off. This would be what Seacon squared of X minus one times attention to the n minus two x dx. So I got the interval tended to the end of X t X. Now this could be split into two different intervals. So this gives me the integral Tange. It'd in power of X. T X is equal to the integral seeking squared of X times a tangent of in minus two of X dx minus the integral Change it in minus two of X dx. Now often the case with integration if you can get the inte grand to show you a function and the derivative substitution is will typically work. So what if z were equal to the tangent of X, then DZ would be the seeking squared of X dx. So I can write this as the integral off. So this looks like, um z to the n minus two DZ and then minus the integral changes in minus two of X t X. And so now this just becomes polynomial integration. How do you integrate Z to the N minus two? That's going to be easy to the N minus one. Divided by N minus one minus this tangent integral. And remember substitution Z was equal to the tangent of X. So this is the changing of X divided by in minus one minus. The integral tended to the n minus two of X dx. And that is what we were after. Sorry, it's tended to the sorry in minus one. So I missed that. So tended to the in minus one of X, divided by n minus one minus the integral. So this is our reduction formula for the integral change it'd in power of X dx. And so the key there was breaking this up in detention square times, changing in minus two, using the Pythagorean identity of seeking, squared and then integrating based on knowing the interval of the derivative change it is seeking squared, which makes that easier in a girl to evaluate.
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