Scribe Post 6
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Transcript of Scribe Post 6
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Here we can apply substitution where u = ln(z). But sometimes, as in this case, it might be necessary to solve for z in terms of u, depending on the complexity of the integrand.
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The variables u and du can now be substituted into the integrand, but now the z must be converted into u so antidifferentiation can occur.
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Since z = e , substituting this into the integrand yields an expression only in terms of u.
u
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Now integration by parts can be applied to the variable u. Don't forget to use LIATE.
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Here's LIATE again in case anyone has forgotten. Remember, it's purpose is to help determine f.
L - Logarithmic (ex. log(x))I - Inverse Trig. (ex. arccos(x))A - Algebraic (ex. x2)T - Trigonometry (ex. cos(x))E - Exponential (ex. ex)
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After applying integration by parts, it's obvious that it's still not simple enough to solve, so we must use parts again.
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Now the full expression can be simplified and eu can easily be antidifferentiatied.
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Remember, u = ln(z), so once the full expression is simplified you must resubstitute ln(z) in. Now the final answer is known.
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This approach involves u-substitution as the first step, where u = x3.
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We can take out the e as a coefficient for the antiderivative, but notice how u has been substituted into the integrand twice where x3 is present.
Two substitutions.
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Now that the the integrand is only in terms of u, we can now antidifferentiate using parts.
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Once again using LIATE, f and g' can be determined for the parts.
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All that is left is to finish antidifferentiating f'g. Also, don't forget that e/3 is to be distributed to the whole antiderivative, thus the brackets.
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Now that there are no indefinite integrals left in the expression, all that is left is some algebraic massage. In this step, I have already factored out 2/3.
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You can factor out a (u + 1)3/2 to further simplify the expression. Don't forget that resubstituting into u is now required.
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Here is the final answer, though further algebraic massage is possible, the answer will still remain the same function.
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LIATE indicates that algebraic functions have priority over trigonometric functions for f, as used here.
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The antiderivative of tan(x) must now be determined, which isn't a fundamental antiderivative.
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But don't forget that we always have our trigonometric functions.
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Define the function for substitution.
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Applying the variable u to the integrand works nicely.
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Don't forget to distribute that negative sign and also to subsitute u = cos(x) once you antidifferentiate.
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Since the antiderivative of tan(x) has now been determined, the full antiderivative of xsec2(x) can easily be determine as above.
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Well, that's it for thursday's scribe.
I hope that anyone who didn't understand what was going on during thursday's class when Mr. K was going through these tougher questions found my dissection helpful. Goodbye for now!
The next scribe is: MrSiwWy! =D