Exercise 4: In this problem, we solve the general case of sec" (x) dx. Note we know the cases n = 1, 2, 3, so we may as well assume n > 3 if this helps. (a) First, suppose n is even. Re-write sec" (x) = sec"-2 (x) · sec2 (x), use the Pythagorean theorem, and a u-substitution of u = tan(x) to solve the integral, similar to our previous methods. (b) Now, suppose n is odd. Use integration by parts in a very similar manner to the sec3 (x) case done in class to show the following: | sec" (x) dx sec"-2 (x) tan(x) – (n – 2) | sec"-2(x) tan² (x) dx. п-2 Conclude that: sec" (2) da = (see"- (r) tan(2) + (n – 2) 2(x) tan(x) + (n – 2) sec dx and hence we are done by induction.
Exercise 4: In this problem, we solve the general case of sec" (x) dx. Note we know the cases n = 1, 2, 3, so we may as well assume n > 3 if this helps. (a) First, suppose n is even. Re-write sec" (x) = sec"-2 (x) · sec2 (x), use the Pythagorean theorem, and a u-substitution of u = tan(x) to solve the integral, similar to our previous methods. (b) Now, suppose n is odd. Use integration by parts in a very similar manner to the sec3 (x) case done in class to show the following: | sec" (x) dx sec"-2 (x) tan(x) – (n – 2) | sec"-2(x) tan² (x) dx. п-2 Conclude that: sec" (2) da = (see"- (r) tan(2) + (n – 2) 2(x) tan(x) + (n – 2) sec dx and hence we are done by induction.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Transcribed Image Text:Exercise 4: In this problem, we solve the general case of sec" (x) dx. Note we know the cases n = 1, 2, 3, so
we may as well assume n > 3 if this helps.
(a) First, suppose n is even. Re-write sec" (x)
= sec"-2 (x) · sec2 (x), use the Pythagorean theorem, and
a u-substitution of u =
tan(x) to solve the integral, similar to our previous methods.
(b) Now, suppose n is odd. Use integration by parts in a very similar manner to the sec3 (x) case done
in class to show the following:
| sec" (x) dx
sec"-2 (x) tan(x) – (n – 2) | sec"-2(x) tan² (x) dx.
п-2
Conclude that:
sec" (2) da = (see"- (r) tan(2) + (n – 2)
2(x) tan(x) + (n – 2)
sec
dx
and hence we are done by induction.
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