2 Answers

Evaluate ∫ sin 2x dx

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MATHS

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

= 1/2sin 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   

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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 /

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