2021 H2 Mathematics Paper 2 Question 5

Integration Techniques

Answers

(a)

15tan⁡5x−x+C{\frac{1}{5} \tan 5x - x + C}

(b)

12sin⁡b−110sin⁡5b{\frac{1}{2} \sin b - \frac{1}{10} \sin 5b}

(c)

ln⁡ln⁡bln⁡a{\ln \frac{\ln b}{\ln a}}

(d)

−14(1+e2x)2+C′{-\frac{1}{4(1+\mathrm{e}^{2x})^2} + C'}

Full solutions

(a)

∫tan⁡25x  dx=∫sec⁡25x−1  dx=15tan⁡5x−x+C  ■\begin{align*} & \int \tan^2 5x \; \mathrm{d}x \\ & = \int \sec^2 5x - 1 \; \mathrm{d}x \\ & = \frac{1}{5} \tan 5x - x + C \; \blacksquare\\ \end{align*}

(b)

∫0bsin⁡2xsin⁡3x  dx=−12∫0bcos⁡5x−cos⁡x  dx=−12[15sin⁡5x−sin⁡x]0b=12sin⁡b−110sin⁡5b  ■\begin{align*} & \int_0^b \sin 2x \sin 3x \; \mathrm{d}x \\ & = -\frac{1}{2} \int_0^b \cos 5x - \cos x \; \mathrm{d}x \\ & = -\frac{1}{2} \left[ \frac{1}{5}\sin 5x - \sin x \right]_0^b \\ & = \frac{1}{2} \sin b - \frac{1}{10} \sin 5b \; \blacksquare\\ \end{align*}

(c)

∫ab1xln⁡x  dx=∫ab1xln⁡x  dx=[ln⁡∣ln⁡x∣]ab=ln⁡ln⁡b−ln⁡ln⁡a=ln⁡ln⁡bln⁡a  ■\begin{align*} & \int_a^b \frac{1}{x\ln x} \; \mathrm{d}x \\ & = \int_a^b \frac{\frac{1}{x}}{\ln x} \; \mathrm{d}x \\ & = \Big[ \ln \left| \ln x \right| \Big]_a^b \\ & = \ln \ln b - \ln \ln a \\ & = \ln \frac{\ln b}{\ln a} \; \blacksquare \end{align*}

(d)

dudx=2e2x\frac{\mathrm{d}u}{\mathrm{d}x} = 2 \mathrm{e}^{2x}
∫e2x(1+e2x)3  dx=∫e2xu312e2x  du=12∫1u3  du=−14u2+C′=−14(1+e2x)2+C′  ■\begin{align*} & \int \frac{\mathrm{e}^{2x}}{(1+\mathrm{e}^{2x})^3} \; \mathrm{d}x \\ & = \int \frac{\mathrm{e}^{2x}}{u^3} \frac{1}{2 \mathrm{e}^{2x}} \; \mathrm{d}u \\ & = \frac{1}{2} \int \frac{1}{u^3} \; \mathrm{d}u \\ & = -\frac{1}{4u^2} + C' \\ & = -\frac{1}{4(1+\mathrm{e}^{2x})^2} + C' \; \blacksquare \end{align*}