Integration techniques: YIJC 2024 Promo Q9

(a) $$\begin{aligned} \int\sin^2 3\theta\,\,\mathrm{d}\theta &=\int\frac{1-\cos6\theta}{2}\,\,\mathrm{d}\theta\\ &=\frac12\theta-\frac1{12}\sin6\theta+C. \end{aligned}$$

9(b) $$\begin{aligned} \int x(\ln x)^2\,\,\mathrm{d} x &=\frac{x^2}{2}(\ln x)^2-\int\frac{x^2}{2}\cdot\frac{2\ln x}{x}\,\,\mathrm{d} x\\ &=\frac12(x\ln x)^2-\int x\ln x\,\,\mathrm{d} x\\ &=\frac12(x\ln x)^2-\left(\frac{x^2}{2}\ln x-\int\frac{x^2}{2}\cdot\frac1x\,\,\mathrm{d} x\right)\\ &=\frac12(x\ln x)^2-\frac12x^2\ln x+\frac{x^2}{4}+C. \end{aligned}$$

9(c) $$\begin{aligned} \int_1^3\frac{x^2}{\sqrt{2x^3-1}}\,\,\mathrm{d} x &=\frac16\int_1^3 6x^2(2x^3-1)^{-1/2}\,\,\mathrm{d} x\\ &=\frac16\left[\frac{(2x^3-1)^{1/2}}{1/2}\right]_1^3\\ &=\frac13\left[(2(3)^3-1)^{1/2}-(2(1)^3-1)^{1/2}\right]\\ &=\frac13(\sqrt{53}-1). \end{aligned}$$

9(d)$\dfrac{\,\mathrm{d} u}{\,\mathrm{d} x}=\mathrm{e}^x=u$.

When $x=0$, $u=1$.

When $x=\ln\sqrt3$, $u=\sqrt3$. $$\begin{aligned} \int_0^{\ln\sqrt3}\frac{\mathrm{e}^{3x}}{\mathrm{e}^{2x}+1}\,\,\mathrm{d} x &=\int_1^{\sqrt3}\frac{u^3}{u^2+1}\cdot\frac1u\,\,\mathrm{d} u\\ &=\int_1^{\sqrt3}\left(1-\frac1{u^2+1}\right)\,\,\mathrm{d} u\\ &=[u-\tan^{-1}u]_1^{\sqrt3}\\ &=(\sqrt3-\tan^{-1}\sqrt3)-(1-\tan^{-1}1)\\ &=\sqrt3-\frac\pi3-1+\frac\pi4\\ &=\sqrt3-1-\frac\pi{12}. \end{aligned}$$