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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