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

<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

Integration by Substitution

A Definite Trigonometric Integral

Techniques of Integration

<p>It is given that</p>
<p style="text-align: center">\[\mathrm{f}(x) = \begin{cases} \tan \left( \frac{\pi}{4a} x \right) & \text{for}\hspace{0.5em} -a &lt; x \le a, \\ \sin \left( \frac{\pi}{2a} x \right) & \text{for}\hspace{0.5em} a &lt; x \le 3a, \end{cases}\]</p>
<p>and that&nbsp;$\mathrm{f}(x) = \mathrm{f}(x - 4a)$ for all real values of $x$, where&nbsp;$a$ is a positive real constant.</p>
<ol class="abc-list"><li data-list-item-id="imported-item-block-3-item-0" data-list-item-path="imported-item-block-3-item-0"><div class="list-item-paragraph">Sketch the graph of $y = \mathrm{f}(x)$ for&nbsp;$-4a \le x \le 7a$.<span class="inline-marks">[3]</span></div></li><li data-list-item-id="imported-item-block-3-item-1" data-list-item-path="imported-item-block-3-item-1"><div class="list-item-paragraph"><p>By using the sketch in part <strong>(a)</strong>, write down an equation relating $\mathrm{f}(x)$ and $\mathrm{f}(-x)$.</p><span class="inline-marks">[1]</span></div></li><li data-list-item-id="imported-item-block-3-item-2" data-list-item-path="imported-item-block-3-item-2"><div class="list-item-paragraph">Show that&nbsp;$\int_{-2a}^{4a} \mathrm{f}(x) \,\mathrm{d}x = \frac{ka}{\pi} (1 + \ln m)$, where&nbsp;$k$ and $m$ are constants to be determined.<span class="inline-marks">[5]</span></div></li></ol>

\[\mathrm{f}(x) = \begin{cases} \tan \left( \frac{\pi}{4a} x \right) & \text{for } -a < x \le a, \\ \sin \left( \frac{\pi}{2a} x \right) & \text{for } a < x \le 3a, \end{cases}\]

and that $\mathrm{f}(x) = \mathrm{f}(x - 4a)$ for all real values of $x$, where $a$ is a positive real constant.

Sketch the graph of $y = \mathrm{f}(x)$ for $-4a \le x \le 7a$.

Show that $\int_{-2a}^{4a} \mathrm{f}(x) \,\mathrm{d}x = \frac{ka}{\pi} (1 + \ln m)$, where $k$ and $m$ are constants to be determined.

Piecewise-Defined Functions

Periodic Functions

Odd and Even Functions

2026 HCI P1 Q7 [Modified]

Functions

<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

Integrating Powers of Sine and Cosine

Area Enclosed by a Curve and the y-axis

Riemann Sums

Sigma Notation

2024 YIJC P1 Q2

Sequence and Series

Applications of Integration

<p>Integrate the following expressions with respect to $x$:</p>
<ol class="abc-list"><li data-list-item-id="item-1775226602625-gv88fi5w8" data-list-item-path="item-1775226602625-gv88fi5w8"><div class="list-item-paragraph">$\displaystyle\displaystyle \displaystyle\int 4\cos{(3x+1)}+7\sec^2{(2x-1)}\, \mathrm{d}x$</div></li><li data-list-item-id="item-1775226603518-uu3fzx7fd" data-list-item-path="item-1775226603518-uu3fzx7fd"><div class="list-item-paragraph">$\displaystyle\displaystyle \displaystyle\int \sin{(5-2x)}-3\sec^2{(6x+2)}\, \mathrm{d}x$</div></li><li data-list-item-id="item-1775226603936-4aj0m9lq8" data-list-item-path="item-1775226603936-4aj0m9lq8"><div class="list-item-paragraph">$\displaystyle\displaystyle \displaystyle\int 10\sec^2{5x}+2\sin{(4x-3)}\, \mathrm{d}x$</div></li><li data-list-item-id="item-1775226604497-e11ag80qd" data-list-item-path="item-1775226604497-e11ag80qd"><div class="list-item-paragraph">$\displaystyle\displaystyle \displaystyle\int \sin{(3x+2)}+\sec^2{(7x-4)}-6\cos{2x}\, \mathrm{d}x$</div></li></ol>

Integrate the following expressions with respect to $x$:

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

Standard Integrals with Quadratic Denominators

<ol class="abc-list"><li data-list-item-id="dss-p2-q4-q-0-content-items-0" data-list-item-path="dss-p2-q4-q-0-content-items-0"><div class="list-item-paragraph">Show that&nbsp;$\dfrac{d}{dx}(\cot x)=-\cosec^2 x$.<span class="inline-marks">[4]</span></div></li><li data-list-item-id="dss-p2-q4-q-0-content-items-1" data-list-item-path="dss-p2-q4-q-0-content-items-1"><div class="list-item-paragraph">Hence find the value of $\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} -\cot^2 x\, dx$.<span class="inline-marks">[4]</span></div></li></ol>

Show that $\dfrac{d}{dx}(\cot x)=-\cosec^2 x$.

Hence find the value of $\displaystyle \int_{\frac{\pi}{4}}^{\frac{\pi}{3}} -\cot^2 x\, dx$.

Derivatives of Reciprocal Trigonometric Functions

Definite Integration as the Reverse of Differentiation

2025 DSS PRELIMS P2 Q4

Differentiation

<p>State the derivative of $\tan x$. Hence, find $\displaystyle\int \sec^3 x \, \mathrm{d}x$.<span class="inline-marks">[5]</span></p>

State the derivative of $\tan x$. Hence, find $\displaystyle\displaystyle\int \sec^3 x \, \mathrm{d}x$.

Derivatives of Trigonometric Functions

Integration by Parts

Integration as an Inverse Process of Differentiation

Cyclic Integration by Parts

2025 CJC CA1 Q3

Techniques of Differentiation

<p>Integrate the following expressions with respect to $x$.</p>
<ol class="abc-list"><li data-list-item-id="item-1770456207213-kvbmf61cb" data-list-item-path="item-1770456207213-kvbmf61cb"><div class="list-item-paragraph">${\displaystyle\displaystyle\int 2\pi \sin \left( \frac{2ex+3}{4} \right) \mathrm{d}x}$</div></li><li data-list-item-id="item-1770456207708-wm5uwy771" data-list-item-path="item-1770456207708-wm5uwy771"><div class="list-item-paragraph">${\displaystyle\displaystyle\int\frac{5-37x}{37x} \ \mathrm{d}x}$</div></li><li data-list-item-id="item-1770456208233-8j99a55ge" data-list-item-path="item-1770456208233-8j99a55ge"><div class="list-item-paragraph">${\displaystyle\int(3+x)\frac{7(3+x)+4}{(3+x)} \ \mathrm{d}x}$</div></li></ol>

Integrate the following expressions with respect to $x$.

Integration of Power Functions

Integration of Reciprocal Linear Functions

<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>

<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>

<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">${\displaystyle\displaystyle\int 3\cos 7x + \sec^2 x \ \mathrm{d}x}$</div></li><li data-list-item-id="item-b" data-list-item-path="item-b"><div class="list-item-paragraph">${\displaystyle\displaystyle\int \frac{5}{x^4} - \cos(7x - 4) \ \mathrm{d}x}$</div></li><li data-list-item-id="item-c" data-list-item-path="item-c"><div class="list-item-paragraph">${\displaystyle\displaystyle\int \sin 5x - \sec^2 2x \ \mathrm{d}x}$</div></li></ol>

${\displaystyle\displaystyle\int 3\cos 7x + \sec^2 x \ \mathrm{d}x}$

${\displaystyle\displaystyle\int \frac{5}{x^4} - \cos(7x - 4) \ \mathrm{d}x}$

${\displaystyle\displaystyle\int \sin 5x - \sec^2 2x \ \mathrm{d}x}$

<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.

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.eyJ1cmwiOiJxdWVzdGlvbi1pbWFnZXMvcGFzdGVkLzE3Njg2MjAxOTA5ODUtb3A5YmF0Z3pkMXEucG5nIiwic2NvcGUiOiJkb3dubG9hZCIsImlhdCI6MTc5MjczMTc5NSwiZXhwIjoxNzkyNzMyMDk1fQ.hx77_QwE3iZDO3W5k426GoohDqwu705inQe58GqeVA8" 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 $\pi \,\sin \left( \frac{\pi \,x}{4}\,\,-\,\,\frac{\pi }{4} \right)$ with respect to$x$.</p>

Integrate $\pi \,\sin \left( \frac{\pi \,x}{4}\,\,-\,\,\frac{\pi }{4} \right)$ with respect to$x$.

<p>Show that&nbsp;$\frac{1-\cos 2x}{1+\cos 2x} = \sec^2 x - 1$. Hence, find $\displaystyle\displaystyle \displaystyle\int \frac{4-4\cos 2x}{1+\cos 2x} \, \mathrm{d}x$.</p>

Show that $\frac{1-\cos 2x}{1+\cos 2x} = \sec^2 x - 1$. Hence, find $\displaystyle\displaystyle \displaystyle\int \frac{4-4\cos 2x}{1+\cos 2x} \, \mathrm{d}x$.

Proving and Simplifying Trigonometric Identities

Trigonometry

<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 $x$.

<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><div class="list-item-paragraph">$\frac{\tan^2 2x(4\sin^2 2x-1)}{\sin^2 2x}$</div></li><li><div class="list-item-paragraph">$\sin\left(\frac{\pi+x}{4}\right)-4x^3+5$</div></li><li><div class="list-item-paragraph">$\pi \cos \left( \frac{\pi x-1}{3} \right)$</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>

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

Math Question Bank