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Section 6 .1, problem 38.
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Here we're asked to derive the reduction formula for how to integrate the secant to the nth power.
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And so they give us the formula that we're asked to derive here.
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So let's start out with looking at the integral of secant to the nth power of x dx.
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Let's just write this as the integral of secant squared of x times the secant of n minus 2 of x.
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Dx.
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Now let's do integration by parts.
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So let's let you equal the secant to the n minus two power of x.
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Then du is going to be n minus two times the secant to the n minus three of x times the derivative of the secant of the secant of the secant of x, which is just secant of x, tangent of x.
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So all of that just using the chain roll.
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So we've got u, d, u, d, v is going to be secant squared of x, dx, therefore v is going to be the tangent of x.
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So use of this to integrate by parts.
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So i come up with the integral of the secant to the nth power of x, dx, is going to be uv...