BOOMERANG INTEGRAL

The boomerang integral: so called because, it comes back to you,
generally occurs when we integrate exponential functions combined
with either sine, cosine, or other trig functions (such as sec 3 x),
which don't fit any of the categories for trig integrals.

Here are 2 examples:

Ex 1

Let's name the integral I.
since we have a product, we will use integration by parts.

 I = e 2x sin x - 2 u = e 2x dv = cos x dx still a product so we use parts again du = 2e 2x dx v = sin x I = e 2x sin x - 2 [- e 2x cos x + 2 ] u 1 = e 2x dv 1 = sin x dx I = e 2x sin x + 2e 2x cos x - 4I du 1 = 2e 2x dx v 1 = - cos x 5I = e 2x sin x + 2e 2x cos x I = (e 2x sin x + 2e 2x cos x) + C

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Ex 2

 I = u = sec x dv = sec 2 x dx du = sec x tan x dx v = tan x I = sec x tan x - using tan 2 x = sec 2 x - 1 I = sec x tan x - I = sec x tan x - I = sec x tan x - I + 2I = sec x tan x + ln | sec x + tan x | u I = ½ (sec x tan x + ln | sec x + tan x |) + C

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This is classic exam question.

Practice

Integrate

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Solution

Integrate

 I = -½ e 3x cos 2x + u = e 3x dv = sin 2x dx still a product so we use parts again du = 3e 3x dx v = - ½ cos 2x I = -½ e 3x cos 2x + u 1 = e 3x dv 1 = cos 2x dx du 1 = 3e 3x dx v 1 = ½ sin 2x so

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