Question

Pls solve this
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Answer 1

Answer:

-1/3

Explanation:

Given: ∫cos 3x dx

Let 3x = u

⇒ x = (u/3)

⇒ dx = (du/3)

⇒ dx = (1/3) du

Now,

Given Question becomes:

⇒ ∫(1/3)cos(u)du = (1/3)sin(u)du

                           = 1/3 sin(3x) + c

Apply the limits:

(i) x → π/6

⇒ (1/3) sin(3 * π/6)

⇒ (1/3) sin(π/2)

⇒ (1/3) * 1

⇒ (1/3)

(ii)

x → (π/3)

⇒ (1/3) sin(3 * π/3)

⇒ (1/3) sin(π)

⇒ 0.

Now,

⇒ π/3 - π/6

⇒ 0 - (1/3)

= -1/3

Hope it helps!

Answer 2

(i) x → π/6

⇒ (1/3) sin(3 * π/6)

⇒ (1/3) sin(π/2)

⇒ (1/3) * 1

⇒ (1/3)

(ii)

x → (π/3)

⇒ (1/3) sin(3 * π/3)

⇒ (1/3) sin(π)

⇒ 0.

Now,

⇒ π/3 - π/6

⇒ 0 - (1/3)

= -1/3