Question

1. which of the following are antiderivatives of ( f(x)=sin x cos x )?

i. ( f(x)=\frac{sin ^{2} x}{2} )
ii. ( f(x)=\frac{cos ^{2} x}{2} )
iii. ( f(x)=\frac{-cos (2 x)}{4} )
(a) i only
(b) ii only
(c) iii only
(d) i and iii only
(e) ii and iii only
calculator allowed

Explanation:

Step1: Differentiate \(F(x)=\frac{\sin^{2}x}{2}\)

Using the chain rule \((u^{n})^\prime = nu^{n - 1}u^\prime\), where \(u=\sin x\), \(n = 2\).
\(F^\prime(x)=\frac{2\sin x\cos x}{2}=\sin x\cos x\)

Step2: Differentiate \(F(x)=\frac{\cos^{2}x}{2}\)

Using the chain rule, where \(u=\cos x\), \(n = 2\).
\(F^\prime(x)=\frac{2\cos x(-\sin x)}{2}=-\sin x\cos x\)

Step3: Differentiate \(F(x)=\frac{-\cos(2x)}{4}\)

Using the chain rule \((\cos(u))^\prime=-\sin(u)u^\prime\), where \(u = 2x\), \(u^\prime=2\).
\(F^\prime(x)=\frac{-(-\sin(2x))\times2}{4}=\frac{2\sin(2x)}{4}\)
Since \(\sin(2x)=2\sin x\cos x\), then \(F^\prime(x)=\frac{2\times2\sin x\cos x}{4}=\sin x\cos x\)

Answer:

D. I and III only