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Evaluate ∫ sin 2x dx

• A. Cos 2x + k
• B. – Cos 2x + k
• C. -1/2Cos 2x + k
• D. 1/2Cos 2x + k

QUICK ANSWER…

C

SOLUTION…  

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:

1. Option A is incorrect.
2. Option B is incorrect.
3. C is correct.
4. D is not the correct answer.

KEY-POINTS…

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 /