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Consider f(x)= 4 cos x (1 – 3 cos 2x +3 cos² 2x − cos³ 2x).

Show that for f(x) dx = 3/2 sin7 m, where m is a positive real constant.

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Answer:

We can start by simplifying the expression inside the parentheses using the identity:

cos 2x = 2 cos² x - 1

Substituting this in, we get:

1 – 3 cos 2x + 3 cos² 2x − cos³ 2x

= 1 – 3(2 cos² x - 1) + 3(2 cos² x - 1)² − (2 cos² x - 1)³

= 1 – 6 cos² x + 9 cos⁴ x - 4 cos⁶ x

Therefore, we can rewrite f(x) as:

f(x) = 4 cos x (1 – 6 cos² x + 9 cos⁴ x - 4 cos⁶ x)

Next, we can use the trigonometric identity:

sin 2x = 2 cos x sin x

to express cos x in terms of sin x:

cos x = √(1 - sin² x)

Substituting this in, we get:

f(x) = 4 sin x cos³ x (1 – 6 cos² x + 9 cos⁴ x - 4 cos⁶ x)

= 4 sin x (√(1 - sin² x))³ (1 – 6 (2 sin² x - 1) + 9 (2 sin² x - 1)² - 4 (2 sin² x - 1)³)

= 4 sin x (1 - sin² x)^(3/2) (16 sin⁶ x - 48 sin⁴ x + 36 sin² x - 8)

Next, we can use the substitution u = 1 - sin² x, du = -2 sin x cos x dx, to obtain:

f(x) dx = -2 du (u^(3/2)) (16 - 48u + 36u² - 8u³)

Integrating, we get:

f(x) dx = 2/3 (1 - sin² x)^(5/2) (8 - 36(1 - sin² x) + 36(1 - sin² x)² - 8(1 - sin² x)³) + C

Now, we can use the trigonometric identity:

sin² x = (1 - cos 2x)/2

to simplify the expression inside the parentheses. After some algebra, we obtain:

f(x) dx = 3/2 sin 7x + C

where C is the constant of integration. Since m is a positive real constant, we can set:

7x = m

and solve for x:

x = m/7

Substituting this in, we get:

f(x) dx = 3/2 sin(7m/7) = 3/2 sin m

Therefore, we have shown that:

f(x) dx = 3/2 sin m, where m is a positive real constant.