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2022 NJC Promo Q6 (i) Use the substitution $x=mtan t$ to find $int{frac{1}{sqrt{{{m}^{2}}+{{x}^{2}}}}text{d}x}$, where $m$ is a positive constant and $0le t<frac{pi }{2}$. [4] (ii) Find $int{frac{x}{sqrt{{{m}^{2}}+{{x}^{2}}}}text{d}x}$. [2] (iii) Hence,
2018 TJC Promo Q5 (b) Without using a graphing calculator, show that $int_{0}^{1}{frac{{{x}^{3}}}{1+{{x}^{2}}}}text{ d}x=kleft( 1-ln 2 right)$, where $k$ is a real number to be determined. [3] Hence find $int_{0}^{1}{{{x}^{2}}{{tan
2022 HCI Promo Q8 (a) Find $int{3t{{tan }^{-1}}left( 3t right)text{d}t}$. [4] (b) Using the substitution $u={{x}^{2}}+1$, show that $int_{0}^{sqrt{7}}{{{x}^{3}}{{left( {{x}^{2}}+1 right)}^{frac{1}{3}}}text{d}x}$ can be expressed as $frac{1}{2}int_{a}^{b}{{{u}^{frac{4}{3}}}-{{u}^{frac{1}{3}}}}text{ d}u$, where $a$ and
2022 EJC Promo Q6 (a) Find $int{x{{text{e}}^{3{{x}^{2}}+1}}}text{d}x$. [1] (b) Find $int{{{sin }^{2}}left( 5x right)},text{d}x$. [3] (c) Find $int{frac{x}{4{{x}^{2}}-4x+17}},text{d}x$. [5] Suggested Video Solutions Suggested Handwritten Solutions (a) (b) (c) (a) (b)
2021 ASRJC P2 Q5 It is given that $R$ is the region bounded by the $x$-axis, the $y$-axis, the line $x=1$ and the curve $y={{x}^{2}}+1$.The area of $R$ can be
2021 YIJC P1 Q1 (a) The diagram shows part of the graph of $y=ln x$ with $n$ rectangles of equal width, where $n$ is a positive integer. (i) Show that
2022 HCI Promo Q9 (i) The diagram shows a sketch of the curve ${{C}_{1}}$ with equation $text{f}left( x right)=x{{text{e}}^{{{x}^{2}}}}$ for $0le xle 1$.The region under ${{C}_{1}}$ is given by $A$.
2022 DHS Promo Q2 (a) Given that $text{f}$ is a continuous and increasing function, explain, with an aid of a sketch, why the value of $underset{nto infty }{mathop{lim }},frac{2}{n}left[ text{f}left(
2021 RI P1 Q8 (i) The curve $G$ has equation $y=frac{1}{1+{{x}^{2}}}$. Sketch the graph of $G$, stating the equation(s) of any asymptote(s) and the coordinates of any turning point(s). [2]
2021 TMJC P1 Q3 Two curves ${{C}_{1}}$ and ${{C}_{2}}$ have equations $y=-sqrt{left( 1-frac{{{left( x-2 right)}^{2}}}{4} right)}$ and $y=-frac{1}{4}{{x}^{2}}$ respectively. (i) Sketch ${{C}_{1}}$ and ${{C}_{2}}$ on the same diagram, stating the