A Math

A Math (IGCSE)

E Math

E Math (IGCSE)

H1 Math

H2 Math

IB AA HL Math

IB AA SL Math

IB AI HL Math

IB AI SL Math

IP Math

N(A) A Math

N(A) E Math

PSLE

PSLE Foundation Mathematics

<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Find $\int \cos^2 (3x)\,\mathrm{d}x$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Find $\int \frac{1-x}{\sqrt{2-x^2}}\,\mathrm{d}x$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">The region bounded by the curve $y = 2\sin x$, the line $y = 3 - 2x$ and the $y$-axis is rotated through $2\pi$ radians about the $y$-axis. Find the volume of the solid obtained, leaving your answer correct to 3 decimal places.<span class="inline-marks">[6]</span></div></li></ol>

Integration by Splitting the Numerator

Integrating Powers of Sine and Cosine

Standard Integrals with Quadratic Denominators

2024 MI PRU2 P1 Q9

Applications of Integration

Techniques of Integration

<p>The diagram below shows a sketch of the graph of $y = \sin^2 x$ for&nbsp;$0 \le x \le \frac{\pi}{2}$.</p>
<p>Rectangles each of width $\frac{\pi}{2n}$ are drawn under the curve for&nbsp;$0 \le x \le \frac{\pi}{2}$.</p>
<div class="graph-block" data-graph-config="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" data-width="400" data-height="300" style="width:400px;height:300px;max-width:100%;display:block;margin:0 auto;"></div>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that&nbsp;$A$, the total area of all the rectangles, is given by $a \sum_{k=1}^{n-1} \sin^2 \frac{k\pi}{2n}$, where&nbsp;$a$ is to be determined.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Find the exact value of $\displaystyle \lim_{n\to\infty} a \sum_{k=1}^{n-1} \sin^2 \frac{k\pi}{2n}$.<span class="inline-marks">[2]</span></div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">Hence, find the value of $\int_{0}^{1} \sin^{-1} \sqrt{y} \ \mathrm{d}y$.<span class="inline-marks">[2]</span></div></li></ol>

The diagram below shows a sketch of the graph of $y = \sin^2 x$ for $0 \le x \le \frac{\pi}{2}$.

Rectangles each of width $\frac{\pi}{2n}$ are drawn under the curve for $0 \le x \le \frac{\pi}{2}$.

Show that $A$, the total area of all the rectangles, is given by $a \sum_{k=1}^{n-1} \sin^2 \frac{k\pi}{2n}$, where $a$ is to be determined.

Limit of an expression

Integrable Forms of Trigonometric Functions

Area Enclosed by a Curve and the y-axis

Riemann Sums

Sigma Notation

2024 YIJC P1 Q2

Sequence and Series

<p>Integrate the following expressions with respect to $x$.</p>
<ol class="abc-list"><li data-list-item-id="item-1770455963381-fwc5am6ib" data-list-item-path="item-1770455963381-fwc5am6ib"><div class="list-item-paragraph">${\displaystyle\displaystyle\int \sin 5x + 3 \sec^{2} 6x \ \mathrm{d}x}$</div></li><li data-list-item-id="item-1770455964311-qvapt4mgu" data-list-item-path="item-1770455964311-qvapt4mgu"><div class="list-item-paragraph">${\displaystyle\displaystyle\int \cos^2 2x + \sin^2 4x \ \mathrm{d}x}$</div></li></ol>

Integrate the following expressions with respect to $x$.

Integration

<p>Integrate the following expressions with respect to $x$.</p>
<ol class="abc-list"><li data-list-item-id="item-1770455648330-yj12ee32x" data-list-item-path="item-1770455648330-yj12ee32x"><div class="list-item-paragraph">${\displaystyle\displaystyle\int \cos^{2} 6x + \tan^{2} 3x \ \mathrm{d}x}$</div></li><li data-list-item-id="item-1770455648877-9h5lkb0sx" data-list-item-path="item-1770455648877-9h5lkb0sx"><div class="list-item-paragraph">${\displaystyle\displaystyle\int \cos^{2} 7x - \sin^{2} 7x \ \mathrm{d}x}$</div></li><li data-list-item-id="item-1770455649410-5y2pgd6da" data-list-item-path="item-1770455649410-5y2pgd6da"><div class="list-item-paragraph">${\displaystyle\displaystyle\int \frac{\sin^{2} 5x + 8}{\cos^{2} 5x} \mathrm{d}x}$</div></li></ol>

<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>
<div class="graph-block" data-graph-config="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" data-width="250" data-height="250" style="width:250px;height:250px;max-width:100%;display:block;margin:0 auto;"></div>
<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.

Integration by Substitution

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

<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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njg2MjAxOTA5ODUtb3A5YmF0Z3pkMXEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjczMTY5NSwiZXhwIjoxNzkyNzMxOTk1fQ.jUTkh-9lrDMe_lnO8mSIJ-maDTkjSag0NqEs1xPpgKs" 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

<p>Integrate the following expressions with respect to $x$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">$\displaystyle\int \sin^2 4x + \sqrt{5x} \, \mathrm{d}x$</div></li><li><div class="list-item-paragraph">$\displaystyle\int \cos^2 (3x + 1) - \cos\ (3x+1)\,\mathrm{d}x$</div></li><li><div class="list-item-paragraph">$\displaystyle\int \tan^2 (1 - 2x) + \sqrt{1-2x}\,\mathrm{d}x$</div></li></ol>

Integrate the following expressions with respect to $x$.

Integration of Power Functions

<p>Integrate the following expressions with respect to $x$.</p>
<ol class="abc-list"><li><div class="list-item-paragraph">$\displaystyle\int \cos^2 (5x) - 1\, \mathrm{d}x$</div></li><li><div class="list-item-paragraph">$\int2 - \sin^2 (3x) \, \mathrm{d}x$</div></li><li><div class="list-item-paragraph">$\displaystyle\int \tan^2 (3x) \, \mathrm{d}x$</div></li></ol>

<p>Integrate the following expressions with respect to $x$.</p>
<ol class="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph"><p>$\pi \sin \left( \frac{2 + \mathrm{e}x}{3} \right)$</p></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">$\frac{3-\sin^2 2x}{\cos^2 2x}$</div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">$\frac{\tan^2(1-x)[3\sin^2(1-x)-1]}{\sin^2(1-x)}$</div></li></ol>

\[\frac{\tan^2(1-x)[3\sin^2(1-x)-1]}{\sin^2(1-x)}\]

<p>Integrate the following expressions with respect to ${x}$</p>
<ol class="abc-list"><li data-list-item-id="item-a"><div class="list-item-content"><span class="list-item-text">$3\sec^2 4x + \sin 3x$</span></div></li><li data-list-item-id="item-b"><div class="list-item-content"><span class="list-item-text">$6 \cos 2x - \sec^2 2x + \sqrt{x}$</span></div></li><li data-list-item-id="item-c"><div class="list-item-paragraph">$\sin^2 2x + \cos^2 x$</div></li></ol>

Integrate the following expressions with respect to ${x}$

<p>Integrate the following expressions with respect to $x$:</p>
<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">$\cos^2 4x + \tan^2 5x$</span></div></li><li><div class="list-item-content"><span class="list-item-text">$\cos^2 5x - \sin^2 5x$</span></div></li><li><div class="list-item-content"><span class="list-item-text">$\frac{\sin^2 3x + 5}{\cos^2 3x}$</span></div></li></ol>

Integrate the following expressions with respect to $x$:

<p>Evaluate the following</p>
<div class="content-block content-block--columns"><div class="question-columns question-columns--2" data-layout="2"><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list"><li><div class="list-item-paragraph">$\displaystyle\displaystyle \displaystyle\int_{0}^{\frac{\pi}{4}} 3\cos 2x + 1 \, \mathrm{d}x$</div></li></ol></div><div class="question-column" style="flex: 1; align-self: flex-start;"><ol class="abc-list" start="2"><li><div class="list-item-paragraph">$\displaystyle\displaystyle \displaystyle\int_{0}^{\frac{\pi}{6}} 4\sin 3x + \cos^2 x \, \mathrm{d}x$</div></li></ol></div></div></div>

Definite Integral

<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="abc-list"><li data-list-item-id="item-a" data-list-item-path="item-a"><div class="list-item-paragraph">Show that&nbsp;$\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">Hence, or otherwise, evaluate $\displaystyle \displaystyle\int_0^{\frac{\pi}{6}} \sin 2\theta \cos 3\theta \, \mathrm{d}\theta$ exactly.<span class="inline-marks">[4]</span></div></li></ol>

Show that $\cos 3\theta = 4\cos^3 \theta - 3\cos \theta$.

Hence, or otherwise, evaluate $\displaystyle \displaystyle\int_0^{\frac{\pi}{6}} \sin 2\theta \cos 3\theta \, \mathrm{d}\theta$ exactly.

Trigonometric Integration Using Product-to-Sum Formulae

2025 ASRJC P1 Q2

<p>By means of substitution $x = \tan \theta$, or otherwise, find the exact value of $\displaystyle\int_{0}^{\infty} \frac{x^{2}}{(1 + x^{2})^{2}}   dx$.</p>

By means of substitution $x = \tan \theta$, or otherwise, find the exact value of $\displaystyle\displaystyle \displaystyle\int_{0}^{\infty} \frac{x^{2}}{(1 + x^{2})^{2}}   dx$.

Improper Integrals and Convergence

<p>$\displaystyle\int_{0}^{\frac{1}{2}}{4{{u}^{2}}\sqrt{1-4{{u}^{2}}}\mathrm{d}u}$, let&nbsp;$2u=\cos x$</p>

$\displaystyle\displaystyle \displaystyle\int_{0}^{\frac{1}{2}}{4{{u}^{2}}\sqrt{1-4{{u}^{2}}}\mathrm{d}u}$, let $2u=\cos x$

<p>$\displaystyle\int{{{\sin }^{2}}x}{{\cos }^{3}}x\mathrm{ d}x$</p>

$\displaystyle\displaystyle \displaystyle\int{{{\sin }^{2}}x}{{\cos }^{3}}x\mathrm{ d}x$

<p>$\displaystyle\int{{{\cos }^{3}}3x}\mathrm{ d}x$</p>

$\displaystyle\displaystyle \displaystyle\int{{{\cos }^{3}}3x}\mathrm{ d}x$

<p class="paragraph-header">Odd powers of $\sin nx$, use$u=\cos nx$<br>Odd powers of $\cos nx$, use$u=\sin nx$<br></p>
<p>$\displaystyle\int{{{{\sin }^{3}}2x}\mathrm{ d}x}$</p>

Odd powers of $\sin nx$, use$u=\cos nx$
Odd powers of $\cos nx$, use$u=\sin nx$


$\displaystyle\displaystyle \displaystyle\int{{{{\sin }^{3}}2x}\mathrm{ d}x}$

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