Integral of sec^3x | Calculus 2 Exercises

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We integrate sec^3(x) using integration by parts. First, we write sec^3x as secx * sec^2x, then we make our choices for u and v and proceed using the known integral of secx as well as the Pythagorean identity variant for tan^2x. #apcalculusbc #calculus2

Integrating secant(x) using longer method:

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Great explanation. Step by step I like it!

jbmu
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Wonderful explanation! Just curious, is there a way to solve this by directly splitting sec^3x into secx * sec^2x then immediately substituting tan^2x + 1 as follows?

∫sec^3x dx
= ∫secx*sec^2x dx (splitting the factors)
= ∫secx*(1+tan^2x) dx (applying sec^2x = 1 + tan^2x identity)
= ∫(secx + secx*tan^2x) dx (distributing secx)
= ∫secx dx + ∫secx*tan^2x dx (splitting the integrals)
= ln|secx + tanx| + ∫secx*tan^2x dx (evaluating the first integral)
= ln|secx + tanx| + ∫secx*(sec^2x - 1) dx (applying tan^2x = sec^2x - 1 identity)
= ln|secx + tanx| + ∫(sec^3x - secx) dx (distributing secx)
= ln|secx + tanx| + ∫sec^3x dx - ∫secx dx (splitting the integrals)
= ln|secx + tanx| + ∫sec^3x dx - ln|secx + tanx| (evaluating the last integral)

then bringing back the left side of the equation...

∫sec^3x dx = ln|secx + tanx| + ∫sec^3x dx - ln|secx + tanx|

but then the "ln|secx + tanx|"s cancel out so I'm left with

∫sec^3x dx = ∫sec^3x dx

I just keep getting stuck, I'm trying this approach because if I saw this on a test this would be my first attempt and I wouldn't be able to search the solution but so I just want to know what error I made that I'm getting stuck in the loop to avoid and save test time. Thanks in advance!

bhaveerathod
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what an annoying teaching style. jumps around; does not built up by writing; has distracting talking head in corner; has hard to read white background; keeps refering back to earlier things that breaks the flow. all in all 1/10

tandemcompound