Evaluate ∫ sin 2x dx
- A. Cos 2x + k
- B. – Cos 2x + k
- C. -1/2Cos 2x + k
- D. 1/2Cos 2x + k
The problem ∫ sin 2x dx suggests integration by substitution.
Let u = 2x, therefore, du/dx = 2,
du = 2dx, or
dx = du/2
Substituting 2x for u, and dx for du/2, we have:
∫ sin u × du/2
= 1/2∫ sin u du
Now we know that
∫ cos A dA = sin A + K, and
∫ sin A dA = -cos A + K
Therefore, 1/2 ∫ sin u du = -1/2 cos u + k
But remember, u = 2x, therefore our final answer becomes: = -1/2 cos 2x + k
Now for the right answer to the above question:
- Option A is incorrect.
- Option B is incorrect.
- C is correct.
- D is not the correct answer.
You may please note these/this:
- Integral of cos x = sin x,
- Integral of sin x = -cos x
- Differential of cos x = – sin x
- Differential of sin x = cos x
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/ culled from 2015 JAMB-UTME mathematics past question 40 /