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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:**

**Option A is incorrect.****Option B is incorrect.****C is correct.****D is not the correct answer.**

**KEY-POINTS…**

**You may please note the****se/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 /**