By Paul Dawkins

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It turns out that a trig substitution will work nicely on the second integral and it will be the same as we did when we had square roots in the problem. aspx Calculus II Okay, at this point we’ve got two options for the remaining integral. We can either use the ideas we learned in the section about integrals involving trig integrals or we could use the following formula. ò cos m q dq = 1 m -1 sin q cos m -1 q + cos m - 2 q dq ò m m Let’s use this formula to do the integral. ò cos 4 1 3 q dq = sin q cos3 q + ò cos 2 q dq 4 4 1 3æ1 1 ö = sin q cos3 q + ç sin q cos q + ò cos 0 q dq ÷ 4 4è2 2 ø 1 3 3 = sin q cos3 q + sin q cos q + q 4 8 8 cos 0 q = 1!

To a certain extent the real subject of this section is how to take advantage of known integrals to do integrals that may not look like anything the ones that we do know how to do or are given in a table of integrals. For the most part we’ll be doing this by using substitution to put integrals into a form that we can deal with. However, not all of the integrals will require a substitution. For some integrals all that we need to do is a little rewriting of the integrand to get into a form that we can deal with.

See if a “simple” substitution will work. Look to see if a simple substitution can be used instead of the often more complicated methods from Calculus II. For example consider both if the following integrals. ó x dx ô 2 õ x -1 òx x 2 - 1 dx The first integral can be done with partial fractions and the second could be done with a trig substitution. However, both could also be evaluated using the substitution u = x 2 - 1 and the work involved in the substitution would be significantly less than the work involved in either partial fractions or trig substitution.

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Calculus II by Paul Dawkins


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