Solutions — Integration by Recognition (Ex 11H)

Year 12 Mathematical Methods Unit 3 · Cambridge §11H · Mr Wong

ANSWER KEY

Part A — Chain-rule recognition

1a $f'(x) =$ $\dfrac{3x^2}{x^3+2}$
1b $\int \tfrac{3x^2}{x^3+2}\,dx = \ln(x^3+2)+c$, divide by 3: $\int \tfrac{x^2}{x^3+2}\,dx = \tfrac{1}{3}\ln(x^3+2) + c$
2a $f'(x) =$ $\dfrac{6x}{3x^2+7}$
2b $\int_0^2 \tfrac{x}{3x^2+7}\,dx = \tfrac{1}{6}\bigl[\ln(3x^2+7)\bigr]_0^2 = \tfrac{1}{6}(\ln 19 - \ln 7) =$ $\tfrac{1}{6}\ln\!\left(\tfrac{19}{7}\right)$
3a $f'(x) =$ $2xe^{x^2+1}$
3b $\int 2xe^{x^2+1}\,dx = e^{x^2+1}+c$, divide by 2: $\int xe^{x^2+1}\,dx = \tfrac{1}{2}e^{x^2+1} + c$
4 $f'(x) = 2x\cos(x^2)$. Hence $\int x\cos(x^2)\,dx = \tfrac{1}{2}\sin(x^2) + c$

Part B — Product-rule recognition

5a Product rule: $f'(x) =$ $\sin(3x) + 3x\cos(3x)$
5b $\int 3x\cos(3x)\,dx = x\sin(3x) - \int\sin(3x)\,dx = x\sin(3x) + \tfrac{1}{3}\cos(3x) + c$. Divide by 3: $\int x\cos(3x)\,dx = \tfrac{1}{3}x\sin(3x) + \tfrac{1}{9}\cos(3x) + c$. Evaluate: at $\tfrac{\pi}{6}$ gives $\tfrac{\pi}{18}\sin\tfrac{\pi}{2} + \tfrac{1}{9}\cos\tfrac{\pi}{2} = \tfrac{\pi}{18}$; at $0$ gives $\tfrac{1}{9}$. $\int_0^{\pi/6} x\cos(3x)\,dx = \dfrac{\pi - 2}{18}$
6 $f'(x) = e^x + xe^x$. So $\int (e^x + xe^x)\,dx = xe^x + c$, hence $\int xe^x\,dx = xe^x - e^x + c =$ $(x-1)e^x + c$
7 $f'(x) = 2xe^x + x^2 e^x = (x^2 + 2x)e^x$. Hence $\int (x^2 + 2x)e^x\,dx = x^2 e^x + c$
8 $f'(x) = 10x(x^2+1)^4$. Divide by 10: $\int x(x^2+1)^4\,dx = \tfrac{1}{10}(x^2+1)^5 + c$

Part C — VCAA-style

9a Let $u = x^2+1$, $u' = 2x$. $\tfrac{d}{dx}\sqrt{u} = \tfrac{u'}{2\sqrt{u}} = \tfrac{2x}{2\sqrt{x^2+1}} =$ $\dfrac{x}{\sqrt{x^2+1}} = g(x)$ ✓
9b By part (a), $\int_0^{\sqrt{3}} g(x)\,dx = \bigl[\sqrt{x^2+1}\bigr]_0^{\sqrt{3}} = \sqrt{4} - \sqrt{1} = 2 - 1 =$ $1$
Y12 Methods · §11H · Solutions Page 1 of 1 · Hand back to Mr Wong