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H2 Math

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N(A) E Math

PSLE

PSLE Foundation Mathematics

<p>Using $x=4\sin\theta$, where&nbsp;$0&lt;\theta&lt;\dfrac{\pi}{2}$, find $\displaystyle\int\dfrac{\sqrt{16-x^2}}{x^2}\,\mathrm{d}x$ for&nbsp;$0&lt;x&lt;4$.</p>

Using $x=4\sin\theta$, where $0<\theta<\dfrac{\pi}{2}$, find $\displaystyle\int\dfrac{\sqrt{16-x^2}}{x^2}\,\mathrm{d}x$ for $0<x<4$.

Integration by Substitution

A Supplied Sine Substitution

Techniques of Integration

<p>Evaluate $\displaystyle\int_0^{\pi/4}\cos^4y\sin y\,\mathrm{d}y$.</p>

Evaluate $\displaystyle\int_0^{\pi/4}\cos^4y\sin y\,\mathrm{d}y$.

Integrable Forms of Trigonometric Functions

A Definite Trigonometric Integral

<p>$ \displaystyle\int{x\sqrt{2x-1}\mathrm{d}x}$</p>

$ \displaystyle\int{x\sqrt{2x-1}\mathrm{d}x}$

Integration by Parts

<p>$\int_{-2}^{\alpha-1} \frac{1}{\left(1-2x-x^2\right)^{\frac{3}{2}}}\,\mathrm{d}x\quad \left(x+1=\sqrt{2}\sin\theta,\ 0 &lt; \alpha &lt; 1\right)$</p>

$\int_{-2}^{\alpha-1} \frac{1}{\left(1-2x-x^2\right)^{\frac{3}{2}}}\,\mathrm{d}x\quad \left(x+1=\sqrt{2}\sin\theta,\ 0 < \alpha < 1\right)$

Integration by Completing the Square

HCI Integration Tutorial Q4 (g)

<p>Evaluate $\int \frac{1}{1+\sin^2 x}\,\mathrm{d}x$ by using the substitution $t = \tan x$.</p>

Evaluate $\int \frac{1}{1+\sin^2 x}\,\mathrm{d}x$ by using the substitution $t = \tan x$.

Standard Integrals with Quadratic Denominators

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><ol class="i-list"><li data-list-item-id="item-a-i" data-list-item-path="item-a/item-a-i"><div class="list-item-paragraph">State the derivative of $\mathrm{e}^{\tan x}$.<span class="inline-marks">[1]</span></div></li><li data-list-item-id="item-a-ii" data-list-item-path="item-a/item-a-ii"><div class="list-item-paragraph">Hence, find $\int \mathrm{e}^{\tan x} (\sec^3 x)(\sin x) \,\mathrm{d}x$.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Write down constants $A$ and $B$ such that, for all values of $x$, $1 - 4x = A(3 - 8x) + B$. Hence, find </div><div class="list-item-paragraph"><p>$\int \frac{1 - 4x}{\sqrt{1 - 4x^2 + 3x}} \,\mathrm{d}x.$</p><span class="inline-marks">[5]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Use the substitution $u = \sqrt{1 + x}$ to find the exact value of $\int_{1}^{3} \frac{x^2}{\left(\sqrt{1 + x}\right)^3} \,\mathrm{d}x$.<span class="inline-marks">[4]</span></div></li></ol>

Derivatives of Trigonometric Functions

Integration as an Inverse Process of Differentiation

Derivatives of Exponential Functions

Chain Rule

2024 VJC Promo Q10

Techniques of Differentiation

<p>Find $\int x^3 \sqrt{x^2 + 1} \mathrm{d}x$.</p>

Find $\int x^3 \sqrt{x^2 + 1} \mathrm{d}x$.

Integration

<p>\[\int \frac{1}{\sin^3x+\cos^3x} \, \mathrm{d}x\]</p>

\[\int \frac{1}{\sin^3x+\cos^3x} \, \mathrm{d}x\]

Integration Using Partial Fractions

<ol class="abc-list"><li data-list-item-id="imported-item-block-1-item-0" data-list-item-path="imported-item-block-1-item-0"><ol class="i-list"><li data-list-item-id="imported-item-block-1-item-0-sub-0" data-list-item-path="imported-item-block-1-item-0/imported-item-block-1-item-0-sub-0"><div class="list-item-paragraph">Find $\int x^2 \cos n x \,\mathrm{d}x$, where&nbsp;$n$ is a positive integer.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="imported-item-block-1-item-0-sub-1" data-list-item-path="imported-item-block-1-item-0/imported-item-block-1-item-0-sub-1"><div class="list-item-paragraph">Hence find $\int_{\pi}^{2\pi} x^2 \cos n x \,\mathrm{d}x$, giving your answers in the form $a\frac{\pi}{n^2}$, where the possible values of $a$ are to be determined.<span class="inline-marks">[2]</span></div></li></ol></li><li data-list-item-id="imported-item-block-1-item-1" data-list-item-path="imported-item-block-1-item-1"><div class="list-item-paragraph">The region bounded by the curve $y = \frac{x\sqrt{x}}{9-x^2}$, the $x$-axis and the lines $x = 0$ and $x = 2$ is rotated through $2\pi$ radians about the $x$-axis. Use the substitution $u = 9 - x^2$ to find the exact volume of the solid obtained, simplifying your answer.<span class="inline-marks">[5]</span></div></li></ol>

The region bounded by the curve $y = \frac{x\sqrt{x}}{9-x^2}$, the $x$-axis and the lines $x = 0$ and $x = 2$ is rotated through $2\pi$ radians about the $x$-axis. Use the substitution $u = 9 - x^2$ to find the exact volume of the solid obtained, simplifying your answer.

Volume of Revolution about the x-axis

N2016 P2 Q2 

Applications of Integration

<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Use the substitution $x = \sin \theta$ to show that&nbsp;$\displaystyle \displaystyle\int_{0}^{1} \sqrt{1-x^{2}} \, \mathrm{d}x = \frac{\pi}{4}$.</span><span class="inline-marks">[4]</span></div></li></ol>
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<p>The diagram shows the region $R$ bounded by the ellipse with equation $4x^{2} + (y-2)^{2} = 4$ and the line $y = 2x$. The ellipse and the line intersect at the origin and the point $(1, 2)$.</p>
<ol class="abc-list" start="2"><li><div class="list-item-content"><span class="list-item-text">Find the area of $R$, giving your answer in terms of $\pi$.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find the volume of solid generated when&nbsp;$R$ is rotated $2\pi$ radians about the $x$-axis.</span><span class="inline-marks">[2]</span></div></li></ol>

Use the substitution $x = \sin \theta$ to show that $\displaystyle \displaystyle\int_{0}^{1} \sqrt{1-x^{2}} \, \mathrm{d}x = \frac{\pi}{4}$.

The diagram shows the region $R$ bounded by the ellipse with equation $4x^{2} + (y-2)^{2} = 4$ and the line $y = 2x$. The ellipse and the line intersect at the origin and the point $(1, 2)$.

Find the area of $R$, giving your answer in terms of $\pi$.

Find the volume of solid generated when $R$ is rotated $2\pi$ radians about the $x$-axis.

Integrating Powers of Sine and Cosine

2025 YIJC P1 Q11

<ol class="abc-list"><li data-list-item-id="item-a-4" data-list-item-path="item-a-4"><div class="list-item-content"><span class="list-item-text">Use the substitution $x=\sin^2 \theta$, where&nbsp;$0 \leq \theta \leq \frac{\pi}{2}$, to find $\displaystyle \displaystyle\int_0^{\frac{1}{2}} \sqrt{\frac{16x}{1-x}} \, \mathrm{d}x$ exactly.</span><span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-b-4" data-list-item-path="item-b-4"><div class="list-item-content"><span class="list-item-text">Find $\displaystyle \displaystyle\int_a^b \frac{|x-1|}{\sqrt{\frac{1}{2}x^2 - x + 1}} \, \mathrm{d}x$ in terms of $a$ and $b$, where&nbsp;$a &lt; 1 &lt; b$.</span><span class="inline-marks">[4]</span></div></li></ol>

Use the substitution $x=\sin^2 \theta$, where $0 \leq \theta \leq \frac{\pi}{2}$, to find $\displaystyle \displaystyle\int_0^{\frac{1}{2}} \sqrt{\frac{16x}{1-x}} \, \mathrm{d}x$ exactly.

Find $\displaystyle \displaystyle\int_a^b \frac{|x-1|}{\sqrt{\frac{1}{2}x^2 - x + 1}} \, \mathrm{d}x$ in terms of $a$ and $b$, where $a < 1 < b$.

Integration by the Reverse Chain Rule

2025 NYJC P1 Q7

<ol class="abc-list"><li data-list-item-id="item-1781605617444-65w2reoy4"><div class="list-item-paragraph">Find $\displaystyle \displaystyle\int \cos (3 \ln x) \, \mathrm{d}x$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-b"><div class="list-item-content"><span class="list-item-text">Let&nbsp;$I$ be the indefinite integral $\displaystyle \displaystyle\int \frac{\mathrm{P}(x)}{1-\sqrt{x}} \, \mathrm{d}x$, $0 &lt; x &lt; 1$, where&nbsp;$\mathrm{P}(x)$ is a polynomial in $x$.</span></div><ol class="i-list"><li data-list-item-id="item-b-i"><div class="list-item-content"><span class="list-item-text">Find $I$ when&nbsp;$\mathrm{P}(x) = 1 - x$.</span><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b-ii"><div class="list-item-content"><span class="list-item-text">By using the substitution $u = 1 - \sqrt{x}$, find $I$ when&nbsp;$\mathrm{P}(x) = 1$.</span><span class="inline-marks">[3]</span></div></li></ol><div class="list-item-after-content">Hence find $I$ when&nbsp;$\mathrm{P}(x) = x$.<span class="inline-marks">[2]</span></div></li></ol>

Let $I$ be the indefinite integral $\displaystyle \displaystyle\int \frac{\mathrm{P}(x)}{1-\sqrt{x}} \, \mathrm{d}x$, $0 < x < 1$, where $\mathrm{P}(x)$ is a polynomial in $x$.

Cyclic Integration by Parts

2025 NJC P1 Q9

<ol class="i-list"><li data-list-item-id="item-i"><div class="list-item-paragraph">Using the substitution $u = x^2 + 3$, show that\[\displaystyle \displaystyle\int \frac{x^3}{\sqrt{x^2 + 3}} \, \mathrm{d}x = \frac{1}{3}\sqrt{(x^2 + 3)^3} - 3\sqrt{x^2 + 3} + C.\]<span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-ii"><div class="list-item-content"><span class="list-item-text">Hence find the exact value of $\displaystyle \displaystyle\int_{-1}^{3} \left| \frac{3x^3}{\sqrt{x^2 + 3}} \right| \, \mathrm{d}x$, leaving your answer in the form $a + b\sqrt{3}$ where&nbsp;$a$ and $b$ are integers to be determined.</span><span class="inline-marks">[3]</span></div></li></ol>

Hence find the exact value of $\displaystyle \displaystyle\int_{-1}^{3} \left| \frac{3x^3}{\sqrt{x^2 + 3}} \right| \, \mathrm{d}x$, leaving your answer in the form $a + b\sqrt{3}$ where $a$ and $b$ are integers to be determined.

2025 MI P1 Q5

<p> Curve $C$ is a circle with radius $2$ and center at the origin with equation $x^2 + y^2 = 4$. <br>Line $L$ has the equation $\sqrt{3}y = x + 2$. The diagram below shows the shaded region $A$ which is enclosed between $C$ and $L$.</p>
<div class="content-block"><figure class="question-figure aligncenter" style="text-align: center;"><img src="https://vocwkpgapvksartxhhor.supabase.co/storage/v1/object/sign/question-images/pasted/1768620190985-op9batgzd1q.png?token=eyJraWQiOiJiZTM3ZjNhZS05OWM4LTQ5ZGQtYjRiNS1iMTBhYmJhNjY5MWYiLCJhbGciOiJIUzI1NiJ9.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njg2MjAxOTA5ODUtb3A5YmF0Z3pkMXEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjczMTcyMiwiZXhwIjoxNzkyNzMyMDIyfQ.PQ9EPrFssMcnRSisuM9_buUY0-Q3V3Ps50gf3qHSCvQ" style="max-width: min(300px, 100%); height: auto; display: block;" class="aligncenter custom-width" /></figure></div>
<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-content"><span class="list-item-text">Use the substitution $x = 2\sin\theta$ to show that the area of region $A$ can be written in the expression $\displaystyle \displaystyle\int_{b}^{a} 4\cos^2\theta \, \mathrm{d}\theta - c$, where&nbsp;$a$, $b$ and $c$ are exact constants to be determined. Hence evaluate this area exactly.</span><span class="inline-marks">[6]</span></div></li><li data-list-item-id="item-b"><div class="list-item-content"><span class="list-item-text">The region $B$ is bounded by the curve $C$, the line $L$ and the line $x = 2$. Find the volume generated when region $B$ is rotated through $2\pi$ radians about the $y$-axis. Give your answer to $2$ decimal places.</span><span class="inline-marks">[3]</span></div></li></ol>

 Curve $C$ is a circle with radius $2$ and center at the origin with equation $x^2 + y^2 = 4$. 
Line $L$ has the equation $\sqrt{3}y = x + 2$. The diagram below shows the shaded region $A$ which is enclosed between $C$ and $L$.

Use the substitution $x = 2\sin\theta$ to show that the area of region $A$ can be written in the expression $\displaystyle \displaystyle\int_{b}^{a} 4\cos^2\theta \, \mathrm{d}\theta - c$, where $a$, $b$ and $c$ are exact constants to be determined. Hence evaluate this area exactly.

The region $B$ is bounded by the curve $C$, the line $L$ and the line $x = 2$. Find the volume generated when region $B$ is rotated through $2\pi$ radians about the $y$-axis. Give your answer to $2$ decimal places.

2025 ASRJC P1 Q7

<ol class="abc-list"><li><div class="list-item-paragraph">By using the substitution ${x = a\tan\theta}$, show that</div><div class="list-item-paragraph" style="text-align: center">${\displaystyle\int \frac{1}{x\sqrt{x^2 + a^2}} \mathrm{d}x = \frac{1}{a}\ln\frac{x}{\sqrt{x^2 + a^2} + a} + c,}$</div><div class="list-item-paragraph">where ${x &gt; 0}$ and ${a}$ is a positive constant.<span class="inline-marks">[4]</span></div></li><li><div class="list-item-paragraph">At time ${t}$ in a chemical reaction, ${x}$ kg of substance $X$ and ${y}$ kg of substance $Y$ are present. Initially, there is $4$ kg of $X$ and $5$ kg of $Y$.</div><div class="list-item-paragraph">The variables ${x}$ and ${y}$ satisfy the equations ${\frac{\mathrm{d}x}{\mathrm{d}t} = xy}$ and ${\frac{\mathrm{d}y}{\mathrm{d}t} = x^2}$.</div><ol class="i-list"><li><div class="list-item-paragraph">Find $\frac{\mathrm{d}y}{\mathrm{d}x}$ in terms of $x$ and $y$, and by solving this differential equation, show that $y = \sqrt{x^2+9}.$<span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Obtain a differential equation in terms of $x$ and $t$ only and hence find an expression of $t$ in terms of $x$.</span><span class="inline-marks">[4]</span></div></li></ol></li></ol>

Solving Differential Equations by Separation of Variables

Connected Rates of Change

2025 ACJC P2 Q5

Differential Equations

Applications of Differentiation

<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-content"><span class="list-item-text">Find $\displaystyle\displaystyle \displaystyle\int_{0}^{4} \mathrm{e}^{|x-1|} \, \mathrm{d}x$, leaving your answer in exact form.</span><span class="inline-marks">[4]</span></div></li><li data-list-item-id="item-b"><div class="list-item-content"><span class="list-item-text">Find $\displaystyle\displaystyle \displaystyle\int \sin^{2} 3x + \tan^{2} 3x \, \mathrm{d}x$.</span><span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c"><div class="list-item-content"><span class="list-item-text">Use the substitution $x = u^{6}$, where&nbsp;$u &gt; 0$, to find $\displaystyle\displaystyle \displaystyle\int \frac{1}{\sqrt{x} + \sqrt[3]{x}} \, \mathrm{d}x$.</span><span class="inline-marks">[5]</span></div></li></ol>

Find $\displaystyle\displaystyle \displaystyle\int_{0}^{4} \mathrm{e}^{|x-1|} \, \mathrm{d}x$, leaving your answer in exact form.

Find $\displaystyle\displaystyle \displaystyle\int \sin^{2} 3x + \tan^{2} 3x \, \mathrm{d}x$.

Use the substitution $x = u^{6}$, where $u > 0$, to find $\displaystyle\displaystyle \displaystyle\int \frac{1}{\sqrt{x} + \sqrt[3]{x}} \, \mathrm{d}x$.

Integrals involving Modulus Functions

Integrals Involving Long Division

2025 EJC P1 Q10

<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Use the substitution $u^2 = 1 + x$ to find $\int \frac{x}{2(1+x)\sqrt{1+x}} \mathrm{d}x$.</span></div></li><li><div class="list-item-paragraph">Hence find $\int \tan^{-1} \sqrt{x+1} \mathrm{d}x$</div></li></ol>

Use the substitution $u^2 = 1 + x$ to find $\int \frac{x}{2(1+x)\sqrt{1+x}} \mathrm{d}x$.

<p>The region bounded by the axes and the curve $y=\cos x$ from $x=0$ to $x=\frac{1}{2}\pi $ is divided into two parts, of areas ${{A}_{1}}$ and ${{A}_{2}}$, by the curve $y=\sin x$ (see diagram). Prove that&nbsp;${{A}_{2}}=\left( \sqrt{2} \right){{A}_{1}}$.<span class="inline-marks">[6]</span></p>
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<p>The two curves meet at $P$. The line through $P$parallel to the $x$-axis meets the $y$-axis at $Q$. The region $OPQ$, bounded by the arc $OP$ and the lines $PQ$ and $QO$, is rotated through $4$ right angles about the $y$-axis to form a solid of revolution of volume $V$. It is given that</p>
<p style="text-align: center">$C = \pi \int_{0}^{\frac{1}{\sqrt{2}}} \left(\sin^{-1} y\right)^2 \mathrm{d}y.$</p>
<ol class="i-list"><li data-list-item-id="item-1762623400041-6ogb2s7m3" data-list-item-path="item-1762623400041-6ogb2s7m3"><div class="list-item-paragraph"><p>By substituting $u={{\sin }^{-1}}y$, show that&nbsp;$V = \pi \int_{0}^{\frac{1}{4}\pi} u^2 \cos u \, \mathrm{d}u$.</p><span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1762623400625-yfwvhf4i2" data-list-item-path="item-1762623400625-yfwvhf4i2"><div class="list-item-paragraph">Show that&nbsp;$\frac{\mathrm{d}}{\mathrm{d}u}\left( {{u}^{2}}\sin u+2u\cos u-2\sin u \right)={{u}^{2}}\cos u$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-1762623401152-duurqg8o9" data-list-item-path="item-1762623401152-duurqg8o9"><div class="list-item-paragraph">Hence find the exact value of $V$.<span class="inline-marks">[2]</span></div></li></ol>
<p style="text-align: center"></p>

The region bounded by the axes and the curve $y=\cos x$ from $x=0$ to $x=\frac{1}{2}\pi $ is divided into two parts, of areas ${{A}_{1}}$ and ${{A}_{2}}$, by the curve $y=\sin x$ (see diagram). Prove that ${{A}_{2}}=\left( \sqrt{2} \right){{A}_{1}}$.

The two curves meet at $P$. The line through $P$parallel to the $x$-axis meets the $y$-axis at $Q$. The region $OPQ$, bounded by the arc $OP$ and the lines $PQ$ and $QO$, is rotated through $4$ right angles about the $y$-axis to form a solid of revolution of volume $V$. It is given that

$C = \pi \int_{0}^{\frac{1}{\sqrt{2}}} \left(\sin^{-1} y\right)^2 \mathrm{d}y.$

Product Rule

N2003 P1 Q14

<p>Find $\displaystyle\int x \sin^{-1}(x^{2}) dx$.</p>

Find $\displaystyle\displaystyle \displaystyle\int x \sin^{-1}(x^{2}) dx$.

<p>Find $\displaystyle\int \frac{1}{x^{2}\sqrt{4 - x^{2}}} dx$ using the substitution $x = \frac{2}{u}$.</p>

Find $\displaystyle\displaystyle \displaystyle\int \frac{1}{x^{2}\sqrt{4 - x^{2}}} dx$ using the substitution $x = \frac{2}{u}$.

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