smc-1204-50-Diff then Integrate

Calculus, 2ADV C4 EQ-Bank 26

Differentiate \(y=x\, \sin (2 x)\). (2 marks)

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Hence, or otherwise, find \(\displaystyle \int x\, \cos (2 x)\, d x\). (2 marks)

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a. \(\dfrac{d y}{d x}=\sin (2 x)+2 x\, \cos (2 x)\)

b. \(\dfrac{1}{2} x\, \sin (2 x)+\dfrac{1}{4} \cos (2 x)+c\)

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a. \(y=x\, \sin (2 x)\)

\(\dfrac{d y}{d x}=\sin (2 x)+2 x\, \cos (2 x)\)

b. \(\text {Using part (a):}\)

\(\displaystyle\int \sin (2 x)+2 x\, \cos (2 x)\, d x=x\, \sin (2 x)+c\)

\(\displaystyle\int \sin (2 x)\, d x+2 \int x\, \cos (2 x)\, d x=x\, \sin (2 x)+c\)

\(2 \displaystyle \int x\, \cos (2 x)\, d x\)	\(=x\, \sin (2 x)-\displaystyle \int \sin (2 x)\, d x+c\)
	\(=x\, \sin (2 x)+\dfrac{1}{2} \cos (2 x)+c\)
\(\displaystyle\int x\, \cos (2 x)\, d x\)	\(=\dfrac{1}{2} x\, \sin (2 x)+\dfrac{1}{4} \cos (2 x)+c\)

Find the derivative of \(x^{2}\tan\,x\) (2 marks)
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Hence, find \(\displaystyle \int (x\,\tan\,x+1)^2\ dx\) (3 marks)
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a. \(\dfrac{dy}{dx}=2x\,\tan\,x + x^2\,\sec^{2}x\)

b. \(x^{2}\tan^{2}x-\dfrac{x^{3}}{3}+x+C\)

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a. \(y=x^{2}\tan\,x\)

\(\text{By product rule:}\)

\(\dfrac{dy}{dx}=2x\,\tan\,x + x^2 \sec^{2}x\)

b. \(\displaystyle \int (x\,\tan\,x+1)^{2}\,dx\)

\[=\int x^2\tan^{2}x + 2x\,\tan\,x +1\ dx\]

\[=\int x^{2}(\sec^{2}x-1)+2x\,\tan\,x +1\,dx\]

\[=\int 2x\,\tan\,x + x^{2}\sec^{2}x-x^{2}+1\,dx\]

\[=x^{2}\tan\,x-\dfrac{x^{3}}{3}+x+C\]

♦♦ Mean mark (b) 34%.

Differentiate `log_e (cos x)` with respect to `x`. (2 marks)

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Hence, or otherwise, evaluate `int_0^(pi/4) tan x\ dx`. (2 marks)

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a. `-tan x`

b. `-log_e (1/sqrt2)\ \ text(or)\ \ 0.35\ \ text{(2 d.p.)}`

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Differentiate `3 + sin 2x`. (1 mark)

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Hence, or otherwise, find `int (cos2x)/(3 + sin 2x)\ dx`. (2 marks)

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a. `2 cos 2x`

b. `1/2 ln (3 + sin 2x) + C\ \ \ \ \ text{(from part (a))}`

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a. `y= 3 + sin 2x`

`dy/dx= 2 cos 2x`

b. `int (cos 2x)/(3 + sin 2x)\ dx`

`= 1/2 int (2 cos 2x)/(3 + sin 2x)\ dx`

`= 1/2 ln (3 + sin 2x) + C\ \ \ \ \ text{(from part (a))}`