Further Calculus

This topic extends the calculus of A Level Mathematics to more powerful integration techniques, Inverse trigonometric functions, volumes of revolution, arc length, and surface area. These tools are essential for university-level mathematics, physics, and engineering.

Board	Paper	Notes
AQA	Paper 1	Integration by parts, inverse trig integrals, volumes, arc length
Edexcel	FP1, FP2	Parts in FP1; inverse trig, volumes, arc length in FP2
OCR (A)	Paper 1	Parts, inverse trig integrals, volumes
CIE (9231)	P1, P2	Parts and volumes in P1; arc length and surface area in P2

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Theorem. For differentiable functions $u(x)$ and $v(x)$:

$$\boxed{\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx}$$

From the product rule:

$$\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$$

Integrating both sides with respect to $x$:

$$uv = \int u\frac{dv}{dx}\,dx + \int v\frac{du}{dx}\,dx$$

Rearranging:

$$\int u\frac{dv}{dx}\,dx = uv - \int v\frac{du}{dx}\,dx \quad \blacksquare$$

When the integral does not simplify in one step, apply integration by parts repeatedly until it does.

Example. Find $\displaystyle\int x^2 e^x\,dx$.

First application: u = x^2$$dv = e^x\,dx. du = 2x\,dx$$v = e^x.

$$\int x^2 e^x\,dx = x^2 e^x - 2\int x e^x\,dx$$

Second application on $\int x e^x\,dx$: u = x$$dv = e^x\,dx. du = dx$$v = e^x.

$$\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C$$

Therefore:

$$\int x^2 e^x\,dx = x^2 e^x - 2(xe^x - e^x) + C = e^x(x^2 - 2x + 2) + C$$

Example. Find $\displaystyle\int e^{ax}\cos bx\,dx$.

Let $I = \int e^{ax}\cos bx\,dx$. First application: $u = e^{ax}$, $dv = \cos bx\,dx$.

$$du = ae^{ax}\,dx, \quad v = \frac{1}{b}\sin bx$$ $$I = \frac{e^{ax}\sin bx}{b} - \frac{a}{b}\int e^{ax}\sin bx\,dx$$

Second application on $\int e^{ax}\sin bx\,dx$: $u = e^{ax}$, $dv = \sin bx\,dx$.

$$du = ae^{ax}\,dx, \quad v = -\frac{1}{b}\cos bx$$ $$\int e^{ax}\sin bx\,dx = -\frac{e^{ax}\cos bx}{b} + \frac{a}{b}\int e^{ax}\cos bx\,dx = -\frac{e^{ax}\cos bx}{b} + \frac{a}{b}I$$

Substituting back:

$$I = \frac{e^{ax}\sin bx}{b} - \frac{a}{b}\left(-\frac{e^{ax}\cos bx}{b} + \frac{a}{b}I\right)$$ $$I = \frac{e^{ax}\sin bx}{b} + \frac{ae^{ax}\cos bx}{b^2} - \frac{a^2}{b^2}I$$ $$I\left(1 + \frac{a^2}{b^2}\right) = e^{ax}\left(\frac{\sin bx}{b} + \frac{a\cos bx}{b^2}\right)$$ $$\boxed{I = \frac{e^{ax}(a\cos bx + b\sin bx)}{a^2 + b^2} + C}$$

A reduction formula expresses an integral $I_n$ (depending on a parameter $n$) in terms of $I_{n-1}$ or $I_{n-2}$.

Write $I_n = \int_0^{\pi/2}\sin^{n-1}x \cdot \sin x\,dx$.

Let $u = \sin^{n-1}x$, $dv = \sin x\,dx$. Then:

$$du = (n-1)\sin^{n-2}x\cos x\,dx, \quad v = -\cos x$$ $$I_n = \bigl[-\sin^{n-1}x\cos x\bigr]_0^{\pi/2} + (n-1)\int_0^{\pi/2}\sin^{n-2}x\cos^2 x\,dx$$

The boundary term vanishes: at x = 0$$\sin 0 = 0; at x = \pi/2$$\cos(\pi/2) = 0.

Using $\cos^2 x = 1 - \sin^2 x$:

$$I_n = (n-1)\int_0^{\pi/2}\sin^{n-2}x\,dx - (n-1)\int_0^{\pi/2}\sin^n x\,dx$$ $$I_n = (n-1)I_{n-2} - (n-1)I_n$$ $$nI_n = (n-1)I_{n-2}$$ $$\boxed{I_n = \frac{n-1}{n}\,I_{n-2}, \quad n \geq 2}$$

The base cases are $I_0 = \displaystyle\int_0^{\pi/2}1\,dx = \dfrac{\pi}{2}$ and $I_1 = \displaystyle\int_0^{\pi/2}\sin x\,dx = 1$.

Example. Using the reduction formula, $I_4 = \dfrac{3}{4}I_2 = \dfrac{3}{4}\cdot\dfrac{1}{2}I_0 = \dfrac{3}{4}\cdot\dfrac{1}{2}\cdot\dfrac{\pi}{2} = \dfrac{3\pi}{16}$.

Caution: warning $0$ and $\pi/2$. For general limits, the boundary term must be evaluated. Example. Find a reduction formula for $I_n = \displaystyle\int x^n e^x\,dx$.

Let u = x^n$$dv = e^x\,dx. Then du = nx^{n-1}\,dx$$v = e^x.

$$\boxed{I_n = x^n e^x - nI_{n-1}}$$

With $I_0 = e^x + C$.

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When the denominator factorises into distinct linear factors, decompose and integrate each term.

Example. $\displaystyle\int \frac{2x+3}{(x+1)(x+2)}\,dx$.

$$\frac{2x+3}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$$

$2x + 3 = A(x+2) + B(x+1)$. Setting $x = -1$: $1 = A$. Setting $x = -2$: $-1 = -B \implies B = 1$.

$$\int \frac{1}{x+1} + \frac{1}{x+2}\,dx = \ln|x+1| + \ln|x+2| + C = \ln|(x+1)(x+2)| + C$$

When the denominator contains an irreducible quadratic $ax^2 + bx + c$ (discriminant $b^2 - 4ac < 0$), the partial fraction has a linear numerator over the quadratic, leading to $\ln$ and $\arctan$ terms.

Example. $\displaystyle\int \frac{3x + 1}{x^2 + 2x + 5}\,dx$.

Complete the square: $x^2 + 2x + 5 = (x+1)^2 + 4$.

Split the numerator to match the derivative of the denominator:

$$\frac{3x+1}{x^2+2x+5} = \frac{\frac{3}{2}(2x+2) + 1 - 3}{x^2+2x+5} = \frac{3}{2}\cdot\frac{2x+2}{x^2+2x+5} - \frac{2}{(x+1)^2+4}$$ $$\int \frac{3x+1}{x^2+2x+5}\,dx = \frac{3}{2}\ln(x^2+2x+5) - 2\cdot\frac{1}{2}\arctan\!\left(\frac{x+1}{2}\right) + C$$ $$= \frac{3}{2}\ln(x^2+2x+5) - \arctan\!\left(\frac{x+1}{2}\right) + C$$

Example. $\displaystyle\int \frac{1}{x(x-1)^2}\,dx$.

$$\frac{1}{x(x-1)^2} = \frac{A}{x} + \frac{B}{x-1} + \frac{C}{(x-1)^2}$$

$1 = A(x-1)^2 + Bx(x-1) + Cx$.

$x = 0$: $1 = A$. $x = 1$: $1 = C$. $x = 2$: $1 = A + 2B + 2C = 1 + 2B + 2 \implies B = -1$.

$$\int\left(\frac{1}{x} - \frac{1}{x-1} + \frac{1}{(x-1)^2}\right)dx = \ln|x| - \ln|x-1| - \frac{1}{x-1} + C$$

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Theorem. The following integrals hold for $a > 0$:

$$\boxed{\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C}$$ $$\boxed{\int \frac{1}{\sqrt{a^2-x^2}}\,dx = \arcsin\frac{x}{a} + C}$$ $$\boxed{\int \frac{1}{a^2-x^2}\,dx = \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right| + C}$$

Let $x = a\tan\theta$ So $dx = a\sec^2\theta\,d\theta$.

$$\int \frac{1}{a^2 + a^2\tan^2\theta}\cdot a\sec^2\theta\,d\theta = \int \frac{a\sec^2\theta}{a^2\sec^2\theta}\,d\theta = \frac{1}{a}\int 1\,d\theta = \frac{\theta}{a} + C$$

Since $\theta = \arctan(x/a)$:

$$\int \frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + C \quad \blacksquare$$

Let $x = a\sin\theta$ So $dx = a\cos\theta\,d\theta$ and $\sqrt{a^2 - x^2} = a\cos\theta$ (for $|\theta| \leq \pi/2$).

$$\int \frac{a\cos\theta}{a\cos\theta}\,d\theta = \int 1\,d\theta = \theta + C = \arcsin\frac{x}{a} + C \quad \blacksquare$$

By partial fractions:

$$\frac{1}{a^2-x^2} = \frac{1}{(a-x)(a+x)} = \frac{1}{2a}\left(\frac{1}{a-x} + \frac{1}{a+x}\right)$$ $$\int \frac{1}{a^2-x^2}\,dx = \frac{1}{2a}\bigl[-\ln|a-x| + \ln|a+x|\bigr] + C = \frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right| + C \quad \blacksquare$$$$\int \frac{1}{a^2 + (x+b)^2}\,dx = \frac{1}{a}\arctan\frac{x+b}{a} + C$$ $$\int \frac{1}{\sqrt{a^2 - (x+b)^2}}\,dx = \arcsin\frac{x+b}{a} + C$$

These follow directly from the standard forms via the substitution $u = x + b$.

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$$\boxed{\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}, \quad |x| < 1}$$ $$\boxed{\frac{d}{dx}\arccos x = -\frac{1}{\sqrt{1-x^2}}, \quad |x| < 1}$$ $$\boxed{\frac{d}{dx}\arctan x = \frac{1}{1+x^2}}$$

Let $y = \arctan x$. Then $x = \tan y$.

Differentiating implicitly with respect to $x$:

$$1 = \sec^2 y \cdot \frac{dy}{dx}$$ $$\frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y} = \frac{1}{1+x^2} \quad \blacksquare$$

Let $y = \arcsin x$. Then $x = \sin y$.

Differentiating implicitly:

$$1 = \cos y \cdot \frac{dy}{dx}$$

Since $\arcsin x$ has range $[-\pi/2, \pi/2]$ We have $\cos y \geq 0$ So $\cos y = \sqrt{1-\sin^2 y} = \sqrt{1-x^2}$.

$$\frac{dy}{dx} = \frac{1}{\cos y} = \frac{1}{\sqrt{1-x^2}} \quad \blacksquare$$

Example. $\dfrac{d}{dx}\arcsin(3x) = \dfrac{3}{\sqrt{1-9x^2}}$.

Example. $\dfrac{d}{dx}\arctan\!\left(\dfrac{x}{2}\right) = \dfrac{1/2}{1 + x^2/4} = \dfrac{2}{4+x^2}$.

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Definition. The volume generated by rotating the region bounded by $y = f(x)$ The $x$-axis, $x = a$ And $x = b$ about the $x$-axis is:

$$\boxed{V = \pi\int_a^b y^2\,dx = \pi\int_a^b [f(x)]^2\,dx}$$

Definition. The volume generated by rotating the region bounded by $x = g(y)$ The $y$-axis, $y = c$ And $y = d$ about the $y$-axis is:

$$\boxed{V = \pi\int_c^d x^2\,dy = \pi\int_c^d [g(y)]^2\,dy}$$

When a curve is given parametrically by $x = x(t)$, $y = y(t)$:

• Rotation about the $x$-axis: $V = \pi\displaystyle\int_{t_1}^{t_2} y^2\,\frac{dx}{dt}\,dt$
• Rotation about the $y$-axis: $V = \pi\displaystyle\int_{t_1}^{t_2} x^2\,\frac{dy}{dt}\,dt$

Example. Find the volume generated by rotating the curve $y = \sqrt{x}$ from $x = 0$ to $x = 4$ About the $x$-axis.

$$V = \pi\int_0^4 (\sqrt{x})^2\,dx = \pi\int_0^4 x\,dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi$$

Example. The curve $x = 2\cos t$, $y = 2\sin t$ for $0 \leq t \leq \pi$ is rotated about the $x$-axis. Find the volume.

$$V = \pi\int_0^{\pi} (2\sin t)^2 \cdot \frac{dx}{dt}\,dt = \pi\int_0^{\pi} 4\sin^2 t \cdot (-2\sin t)\,dt$$ $$= -8\pi\int_0^{\pi}\sin^3 t\,dt = 8\pi\int_0^{\pi}\sin^3 t\,dt$$

Using $\sin^3 t = \sin t(1-\cos^2 t)$ and the substitution $u = \cos t$:

$$= 8\pi\int_{-1}^{1}(1-u^2)\,du = 8\pi\left[u - \frac{u^3}{3}\right]_{-1}^1 = 8\pi\left(\frac{2}{3} - \left(-\frac{2}{3}\right)\right) = \frac{32\pi}{3}$$

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Theorem. For a curve $y = f(x)$ from $x = a$ to $x = b$:

$$\boxed{s = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx}$$

For a curve given parametrically by $x = x(t)$, $y = y(t)$ from $t = t_1$ to $t = t_2$:

$$\boxed{s = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt}$$

Derivation (Cartesian). The arc length element $ds$ satisfies $ds^2 = dx^2 + dy^2$ by the Pythagorean theorem applied to an infinitesimal segment. Therefore:

$$ds = \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx$$

Integrating from $a$ to $b$ gives the total arc length.

Example. Find the arc length of $y = \ln(\cos x)$ from $x = 0$ to $x = \pi/3$.

$$\frac{dy}{dx} = \frac{-\sin x}{\cos x} = -\tan x$$ $$s = \int_0^{\pi/3}\sqrt{1+\tan^2 x}\,dx = \int_0^{\pi/3}\sec x\,dx = \Bigl[\ln|\sec x + \tan x|\Bigr]_0^{\pi/3}$$ $$= \ln(2 + \sqrt{3}) - \ln(1) = \ln(2+\sqrt{3})$$

Theorem. The surface area generated by rotating $y = f(x)$ from $x = a$ to $x = b$ about the $x$-axis:

$$\boxed{S = 2\pi\int_a^b y\,\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx}$$

For a parametric curve rotated about the $x$-axis:

$$\boxed{S = 2\pi\int_{t_1}^{t_2} y\,\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt}$$

Example. Find the surface area generated by rotating $y = x^2$ from $x = 0$ to $x = 1$ about the $x$-axis.

$$S = 2\pi\int_0^1 x^2\sqrt{1+4x^2}\,dx$$

Let $x = \frac{1}{2}\tan\theta$, $dx = \frac{1}{2}\sec^2\theta\,d\theta$. When $x = 0$ $\theta = 0$; when $x = 1$, $\theta = \arctan 2$.

$$S = 2\pi\int_0^{\arctan 2}\frac{\tan^2\theta}{4}\cdot\sec\theta\cdot\frac{1}{2}\sec^2\theta\,d\theta = \frac{\pi}{4}\int_0^{\arctan 2}\tan^2\theta\sec^3\theta\,d\theta$$

Using $\tan^2\theta = \sec^2\theta - 1$ and integrating by parts with $u = \sec\theta$ $dv = \sec^2\theta\,d\theta$:

This integral evaluates to $\dfrac{\pi}{4}\left[\dfrac{1}{4}\sec\theta\tan\theta + \dfrac{1}{4}\ln|\sec\theta+\tan\theta| - \dfrac{1}{4}\sec\theta\tan\theta + \dfrac{1}{8}\ln|\sec\theta+\tan\theta|\right]_0^{\arctan 2}$.

Simplifying with $\sec(\arctan 2) = \sqrt{5}$ and $\tan(\arctan 2) = 2$:

$$S = \frac{9\pi\sqrt{5}}{16} - \frac{\pi}{32}\ln(2+\sqrt{5})$$

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Integral	Result
$\displaystyle\int\frac{1}{a^2+x^2}\,dx$	$\dfrac{1}{a}\arctan\dfrac{x}{a}+C$
$\displaystyle\int\frac{1}{\sqrt{a^2-x^2}}\,dx$	$\arcsin\dfrac{x}{a}+C$
$\displaystyle\int\frac{1}{a^2-x^2}\,dx$	$\dfrac{1}{2a}\ln\left\|\dfrac{a+x}{a-x}\right\|+C$
$\dfrac{d}{dx}\arcsin x$	$\dfrac{1}{\sqrt{1-x^2}}$
$\dfrac{d}{dx}\arctan x$	$\dfrac{1}{1+x^2}$
Vol. About $x$-axis	$\pi\displaystyle\int_a^b y^2\,dx$
Arc length	$\displaystyle\int\sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$

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Problem 1

Find $\displaystyle\int x^3 e^{-x}\,dx$.

Hint 1

Apply integration by parts three times, reducing the power of $x$ each time.

Answer 1

First: $u = x^3$$dv = e^{-x}\,dx$. $du = 3x^2\,dx$$v = -e^{-x}$.

$\displaystyle\int x^3 e^{-x}\,dx = -x^3 e^{-x} + 3\int x^2 e^{-x}\,dx$

Second: u = x^2$$dv = e^{-x}\,dx. $\int x^2 e^{-x}\,dx = -x^2 e^{-x} + 2\int x e^{-x}\,dx$.

Third: u = x$$dv = e^{-x}\,dx. $\int x e^{-x}\,dx = -xe^{-x} + \int e^{-x}\,dx = -xe^{-x} - e^{-x}$.

Combining: $\int x^3 e^{-x}\,dx = -x^3 e^{-x} - 3x^2 e^{-x} - 6x e^{-x} - 6e^{-x} + C = -e^{-x}(x^3 + 3x^2 + 6x + 6) + C$.

Problem 2

Find a reduction formula for $I_n = \displaystyle\int \cos^n x\,dx$ for $n \geq 2$.

Hint 2

Write $\cos^n x = \cos^{n-1}x \cdot \cos x$ and apply integration by parts with $u = \cos^{n-1}x$$dv = \cos x\,dx$. Use $\sin^2 x = 1 - \cos^2 x$.

Answer 2

$I_n = \int\cos^{n-1}x\cos x\,dx$. Let $u = \cos^{n-1}x$$dv = \cos x\,dx$.

du = -(n-1)\cos^{n-2}x\sin x\,dx$$v = \sin x.

$I_n = \cos^{n-1}x\sin x + (n-1)\int\cos^{n-2}x\sin^2 x\,dx$

$= \cos^{n-1}x\sin x + (n-1)\int\cos^{n-2}x(1-\cos^2 x)\,dx$

$= \cos^{n-1}x\sin x + (n-1)I_{n-2} - (n-1)I_n$

$nI_n = \cos^{n-1}x\sin x + (n-1)I_{n-2}$

$\boxed{I_n = \frac{1}{n}\cos^{n-1}x\sin x + \frac{n-1}{n}I_{n-2} + C}$

Problem 3

Evaluate $\displaystyle\int \frac{2x-1}{x^2+4x+13}\,dx$.

Hint 3

Complete the square: $x^2+4x+13 = (x+2)^2+9$. Split the numerator into a multiple of $(2x+4)$ plus a constant.

Answer 3

$x^2+4x+13 = (x+2)^2+9$. Write $2x-1 = (2x+4) - 5$.

$\displaystyle\int\frac{2x+4}{(x+2)^2+9}\,dx - 5\int\frac{1}{(x+2)^2+9}\,dx$

$= \ln(x^2+4x+13) - \frac{5}{3}\arctan\!\left(\frac{x+2}{3}\right) + C$

Problem 4

Evaluate $\displaystyle\int_0^{\pi/2} \sin^5 x\,dx$ using the reduction formula.

Hint 4

Use $I_n = \dfrac{n-1}{n}I_{n-2}$ with $I_1 = 1$.

Answer 4

$I_5 = \dfrac{4}{5}I_3 = \dfrac{4}{5}\cdot\dfrac{2}{3}I_1 = \dfrac{4}{5}\cdot\dfrac{2}{3}\cdot 1 = \dfrac{8}{15}$.

Problem 5

Find the volume generated by rotating the region bounded by $y = x^2$$y = 0$$x = 0$$x = 2$ about the $y$-axis.

Hint 5

Express $x$ in terms of $y$ ($x = \sqrt{y}$) and integrate with respect to $y$ from $0$ to $4$ Or use the Shell method: $V = 2\pi\int_0^2 x\cdot x^2\,dx$.

Answer 5

Using the disc method about the $y$-axis: $V = \pi\int_0^4 x^2\,dy = \pi\int_0^4 y\,dy = \pi\left[\frac{y^2}{2}\right]_0^4 = 8\pi$.

Problem 6

Find $\dfrac{d}{dx}\left[x\arcsin x + \sqrt{1-x^2}\right]$ and hence evaluate $\displaystyle\int_0^{1/2} \arcsin x\,dx$.

Hint 6

Differentiate using the product rule and the chain rule with $\dfrac{d}{dx}\arcsin x$.

Answer 6

$\dfrac{d}{dx}\bigl[x\arcsin x + \sqrt{1-x^2}\bigr] = \arcsin x + \dfrac{x}{\sqrt{1-x^2}} + \dfrac{-x}{\sqrt{1-x^2}} = \arcsin x$.

Therefore $\displaystyle\int \arcsin x\,dx = x\arcsin x + \sqrt{1-x^2} + C$.

$\displaystyle\int_0^{1/2}\arcsin x\,dx = \left[x\arcsin x + \sqrt{1-x^2}\right]_0^{1/2} = \frac{1}{2}\cdot\frac{\pi}{6} + \frac{\sqrt{3}}{2} - 1 = \frac{\pi}{12} + \frac{\sqrt{3}}{2} - 1$.

Problem 7

Find the arc length of the curve $y = \dfrac{x^3}{6} + \dfrac{1}{2x}$ from $x = 1$ to $x = 3$.

Hint 7

Compute $\dfrac{dy}{dx} = \dfrac{x^2}{2} - \dfrac{1}{2x^2}$. Show that $1+\left(\dfrac{dy}{dx}\right)^2$ is a perfect square.

Answer 7

$\dfrac{dy}{dx} = \dfrac{x^2}{2} - \dfrac{1}{2x^2}$.

$1+\left(\dfrac{dy}{dx}\right)^2 = 1 + \dfrac{x^4}{4} - \dfrac{1}{2} + \dfrac{1}{4x^4} = \dfrac{x^4}{4} + \dfrac{1}{2} + \dfrac{1}{4x^4} = \left(\dfrac{x^2}{2} + \dfrac{1}{2x^2}\right)^2$

$s = \displaystyle\int_1^3\left(\dfrac{x^2}{2} + \dfrac{1}{2x^2}\right)dx = \left[\dfrac{x^3}{6} - \dfrac{1}{2x}\right]_1^3 = \left(\dfrac{9}{2}-\dfrac{1}{6}\right)-\left(\dfrac{1}{6}-\dfrac{1}{2}\right) = \dfrac{14}{3}$.

Problem 8

Evaluate $\displaystyle\int_0^{\infty}\frac{1}{4+x^2}\,dx$.

Hint 8

Use $\displaystyle\int\frac{1}{a^2+x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a}$. Here $a = 2$.

Answer 8

$\displaystyle\int_0^{\infty}\frac{1}{4+x^2}\,dx = \left[\frac{1}{2}\arctan\frac{x}{2}\right]_0^{\infty} = \frac{1}{2}\cdot\frac{\pi}{2} - 0 = \frac{\pi}{4}$.

Problem 9

The curve $x = t^2$$y = t^3$ for $0 \leq t \leq 2$ is rotated about the $x$-axis. Find the volume of Revolution.

Hint 9

Use $V = \pi\displaystyle\int_{t_1}^{t_2}y^2\,\frac{dx}{dt}\,dt$. Here $\dfrac{dx}{dt} = 2t$.

Answer 9

$V = \pi\displaystyle\int_0^2 t^6 \cdot 2t\,dt = 2\pi\int_0^2 t^7\,dt = 2\pi\left[\frac{t^8}{8}\right]_0^2 = 2\pi\cdot\frac{256}{8} = 64\pi$.

Problem 10

Find $\displaystyle\int e^x\sin 2x\,dx$.

Hint 10

Apply integration by parts twice. Keep $u = e^x$ on both applications. The original integral will reappear.

Answer 10

Let $I = \displaystyle\int e^x\sin 2x\,dx$. First: $u = e^x$$dv = \sin 2x\,dx$. $du = e^x\,dx$$v = -\frac{1}{2}\cos 2x$.

$I = -\frac{1}{2}e^x\cos 2x + \frac{1}{2}\int e^x\cos 2x\,dx$.

Second on $\int e^x\cos 2x\,dx$: u = e^x$$dv = \cos 2x\,dx. $du = e^x\,dx$ $v = \frac{1}{2}\sin 2x$.

$\int e^x\cos 2x\,dx = \frac{1}{2}e^x\sin 2x - \frac{1}{2}\int e^x\sin 2x\,dx = \frac{1}{2}e^x\sin 2x - \frac{1}{2}I$.

$I = -\frac{1}{2}e^x\cos 2x + \frac{1}{4}e^x\sin 2x - \frac{1}{4}I$.

$\frac{5}{4}I = e^x\left(\frac{\sin 2x}{4} - \frac{\cos 2x}{2}\right)$.

$\boxed{I = \frac{e^x(\sin 2x - 2\cos 2x)}{5} + C}$

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Problem. If $y = x^2 e^{3x}$ Find $\dfrac{d^4 y}{dx^4}$.

Solution. We use Leibniz’s rule: $(uv)^{(n)} = \displaystyle\sum_{k=0}^{n} \binom{n}{k} u^{(k)} v^{(n-k)}$.

Let $u = x^2$ and $v = e^{3x}$.

• u' = 2x$$u'' = 2$$u''' = 0$$u^{(4)} = 0.
• $v^{(k)} = 3^k e^{3x}$ for all $k$.

$$\frac{d^4 y}{dx^4} = \binom{4}{0} x^2 \cdot 3^4 e^{3x} + \binom{4}{1} 2x \cdot 3^3 e^{3x} + \binom{4}{2} 2 \cdot 3^2 e^{3x} + 0 + 0$$ $$= 81x^2 e^{3x} + 4 \cdot 54x e^{3x} + 6 \cdot 18 e^{3x}$$ $$\boxed{= (81x^2 + 216x + 108)e^{3x}}$$

Problem. Establish and use a reduction formula for $I_n = \int x^n e^x\,dx$.

Solution. Using integration by parts with u = x^n$$dv = e^x\,dx:

$$I_n = x^n e^x - \int nx^{n-1} e^x\,dx = x^n e^x - nI_{n-1}$$

Therefore $\boxed{I_n = x^n e^x - nI_{n-1}}$ with $I_0 = e^x + C$.

To find $I_3$:

I_1 = x e^x - e^x$$I_2 = x^2 e^x - 2x e^x + 2e^x$$I_3 = x^3 e^x - 3x^2 e^x + 6x e^x - 6e^x.

$$\boxed{I_3 = (x^3 - 3x^2 + 6x - 6)e^x + C}$$

Problem. Determine whether $\displaystyle\int_0^1 \frac{1}{\sqrt{x}}\,dx$ converges, And evaluate if it does.

Solution. The integrand is undefined at $x = 0$. Write:

$$\int_0^1 x^{-1/2}\,dx = \lim_{a \to 0^+} \int_a^1 x^{-1/2}\,dx = \lim_{a \to 0^+} \left[2x^{1/2}\right]_a^1 = \lim_{a \to 0^+} (2 - 2\sqrt{a}) = 2$$

Since the limit exists and is finite, the integral converges. $\boxed{\displaystyle\int_0^1 \frac{1}{\sqrt{x}}\,dx = 2}$

Problem. Evaluate $\displaystyle\int_0^{\pi/2} \frac{1}{1 + \sin x}\,dx$ using the Weierstrass substitution.

Solution. Let $t = \tan(x/2)$ So $\sin x = \dfrac{2t}{1+t^2}$ and $dx = \dfrac{2\,dt}{1+t^2}$.

When $x = 0$: $t = 0$. When $x = \pi/2$: $t = 1$.

$$\int_0^1 \frac{1}{1 + \frac{2t}{1+t^2}} \cdot \frac{2\,dt}{1+t^2} = \int_0^1 \frac{2\,dt}{(1+t^2) + 2t} = \int_0^1 \frac{2\,dt}{t^2 + 2t + 1} = \int_0^1 \frac{2\,dt}{(t+1)^2}$$ $$= \left[-\frac{2}{t+1}\right]_0^1 = -1 + 2 = \boxed{1}$$

Problem. A curve is given by $x = t - \sin t$, $y = 1 - \cos t$ for $0 \leq t \leq 2\pi$. Find The total arc length.

Solution. $\dfrac{dx}{dt} = 1 - \cos t$, $\dfrac{dy}{dt} = \sin t$.

$$s = \int_0^{2\pi} \sqrt{(1-\cos t)^2 + \sin^2 t}\,dt = \int_0^{2\pi} \sqrt{1 - 2\cos t + \cos^2 t + \sin^2 t}\,dt$$ $$= \int_0^{2\pi} \sqrt{2 - 2\cos t}\,dt = \int_0^{2\pi} \sqrt{4\sin^2(t/2)}\,dt = \int_0^{2\pi} 2|\sin(t/2)|\,dt$$

For $0 \leq t \leq 2\pi$, $\sin(t/2) \geq 0$ So:

$$s = 2\int_0^{2\pi} \sin(t/2)\,dt = 2\left[-2\cos(t/2)\right]_0^{2\pi} = 2(2 + 2) = \boxed{8}$$

Problem. Evaluate $\displaystyle\lim_{x \to 0} \frac{x - \sin x}{x^3}$.

Solution. Expand $\sin x$ as a Maclaurin series:

$$\sin x = x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots$$ $$\frac{x - \sin x}{x^3} = \frac{x - \left(x - \frac{x^3}{6} + \frac{x^5}{120} - \cdots\right)}{x^3} = \frac{\frac{x^3}{6} - \frac{x^5}{120} + \cdots}{x^3} = \frac{1}{6} - \frac{x^2}{120} + \cdots$$

Taking $x \to 0$: $\boxed{\displaystyle\lim_{x \to 0} \frac{x - \sin x}{x^3} = \frac{1}{6}}$

Problem. Evaluate $\displaystyle\int \arcsin x\,dx$.

Solution. Use integration by parts with $u = \arcsin x$, $dv = dx$:

$$du = \frac{1}{\sqrt{1-x^2}}\,dx, \quad v = x$$ $$\int \arcsin x\,dx = x\arcsin x - \int \frac{x}{\sqrt{1-x^2}}\,dx$$

For the second integral, let $w = 1 - x^2$, $dw = -2x\,dx$:

$$\int \frac{x}{\sqrt{1-x^2}}\,dx = -\sqrt{1-x^2}$$ $$\boxed{\int \arcsin x\,dx = x\arcsin x + \sqrt{1-x^2} + C}$$

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Integration by parts is the reverse of the product rule, allowing you to transfer derivatives between functions. Think of it as a negotiation: you give away one function’s derivative in exchange for integrating the other. Reduction formulae create a staircase where each integral is expressed in terms of a simpler one, allowing you to climb down to a base case you can evaluate directly. Inverse trigonometric functions arise because their derivatives produce rational functions, creating a bridge between algebraic and trigonometric integration. Volumes of revolution are computed by summing infinitesimally thin discs stacked along an axis, like building a solid from a stack of coins. Arc length measures the actual distance along a curve rather than straight-line approximation.

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Pitfall	Correct Approach
Forgetting the chain rule when differentiating composite inverse trig functions	Always write $\dfrac{d}{dx}\!\left[\arcsin(u)\right] = \dfrac{u'}{\sqrt{1-u^2}}$
Using $\ln \| x \|$ before checking if the integral is improper	Check for discontinuities in the interval first
Forgetting $+C$ on every antiderivative	Every indefinite integral needs an arbitrary constant
Applying reduction formulae without checking the base case	Always state $I_0$ or $I_1$ explicitly
Confusing $\dfrac{d^n y}{dx^n}$ notation with $\left(\dfrac{dy}{dx}\right)^n$	$\dfrac{d^n y}{dx^n}$ is the $n$-th derivative, not the $n$-th power

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Using the substitution $u = e^x$ Find $\displaystyle\int \frac{e^x}{e^{2x} + 1}\,dx$.

Solution

$u = e^x$, $du = e^x\,dx$.

$$\int \frac{du}{u^2 + 1} = \arctan u + C = \boxed{\arctan(e^x) + C}$$

The reduction formula $I_n = \displaystyle\int_0^{\pi/4} \tan^n x\,dx$ satisfies $I_n = \dfrac{1}{n-1} - I_{n-2}$ for $n \geq 2$. Given $I_0 = \dfrac{\pi}{4}$ and $I_1 = \dfrac{1}{2}\ln 2$ Find $I_3$.

Solution

$I_3 = \dfrac{1}{2} - I_1 = \dfrac{1}{2} - \dfrac{1}{2}\ln 2 = \dfrac{1}{2}(1 - \ln 2)$.

To verify: $I_2 = \dfrac{1}{1} - I_0 = 1 - \dfrac{\pi}{4}$. Then $I_3 = \dfrac{1}{2} - I_1 = \dfrac{1}{2} - \dfrac{1}{2}\ln 2$. Consistent. $\boxed{I_3 = \dfrac{1}{2}(1 - \ln 2)}$

Find the area enclosed by the curve $x = t^2$, $y = t^3 - t$ for $-1 \leq t \leq 1$.

Solution

Using the parametric area formula $A = \displaystyle\int y\,\frac{dx}{dt}\,dt$:

$$A = \int_{-1}^{1} (t^3 - t)(2t)\,dt = 2\int_{-1}^{1} (t^4 - t^2)\,dt = 2\left[\frac{t^5}{5} - \frac{t^3}{3}\right]_{-1}^{1}$$

Since $t^5 - \frac{5}{3}t^3$ is odd (each term is odd), the integral from $-1$ to $1$ is zero.

$\boxed{A = 0}$ (the curve traces back over itself symmetrically).

Prove that $\dfrac{d}{dx}\!\left[\arctan x\right] = \dfrac{1}{1+x^2}$ from first principles Using implicit differentiation.

Solution

Let $y = \arctan x$ So $x = \tan y$. Differentiating implicitly with respect to $x$:

$$1 = \sec^2 y \cdot \frac{dy}{dx}$$ $$\frac{dy}{dx} = \cos^2 y = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$$

$\blacksquare$

Evaluate $\displaystyle\int_0^1 \frac{\ln x}{1+x}\,dx$ Expressing your answer in terms Of $\displaystyle\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$.

Solution

Expand $\dfrac{1}{1+x} = \displaystyle\sum_{n=0}^{\infty} (-1)^n x^n$ for $|x| < 1$:

$$\int_0^1 \ln x \sum_{n=0}^{\infty} (-1)^n x^n\,dx = \sum_{n=0}^{\infty} (-1)^n \int_0^1 x^n \ln x\,dx$$

Using integration by parts or the standard result $\displaystyle\int_0^1 x^n \ln x\,dx = -\frac{1}{(n+1)^2}$:

$$= -\sum_{n=0}^{\infty} \frac{(-1)^n}{(n+1)^2} = -\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^2} = \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}$$

This equals $-\dfrac{\pi^2}{12}$.

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Integration techniques (substitution, parts, partial fractions) are essential tools for solving Differential equations. See Differential Equations.

Taylor and Maclaurin expansions provide powerful tools for evaluating integrals that cannot be found in closed form. See Maclaurin and Taylor Series.

Arc length and area calculations are used extensively in mechanics for work-energy problems. See Circular Motion.

The inverse hyperbolic functions arise from integration: $\displaystyle\int \frac{dx}{\sqrt{x^2+a^2}} = \operatorname{arsinh}(x/a) + C$. See Hyperbolic Functions.

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When choosing $u$ and $dv$ for integration by parts, use the LIATE priority:

• Logarithmic functions
• Inverse trigonometric functions
• Algebraic functions (polynomials)
• Trigonometric functions
• Exponential functions

The function higher on the list should be chosen as $u$.

For integrals involving rational functions of $\sin x$ and $\cos x$ The substitution $t = \tan(x/2)$ converts them to rational functions of $t$:

\sin x = \dfrac{2t}{1+t^2}$$\cos x = \dfrac{1-t^2}{1+t^2}$$dx = \dfrac{2\,dt}{1+t^2}.

Form	Result
$\displaystyle\int \frac{f'(x)}{f(x)}\,dx$	$\ln \| f(x) \| + C$
$\displaystyle\int \frac{f'(x)}{\sqrt{f(x)}}\,dx$	$2\sqrt{f(x)} + C$
$\displaystyle\int f(x) \cdot f'(x)\,dx$	$\dfrac{[f(x)]^2}{2} + C$

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Result	Formula
Integration by parts	$\displaystyle\int u\,dv = uv - \int v\,du$
Reduction formula (by parts)	Express $I_n$ in terms of $I_{n-1}$ or $I_{n-2}$
Arc length (Cartesian)	$s = \displaystyle\int_a^b \sqrt{1+\left(\frac{dy}{dx}\right)^2}\,dx$
Arc length (parametric)	$s = \displaystyle\int_\alpha^\beta \sqrt{\dot{x}^2+\dot{y}^2}\,dt$
Area under parametric curve	$A = \displaystyle\int y\frac{dx}{dt}\,dt$
Surface of revolution	$S = 2\pi\displaystyle\int_a^b y\sqrt{1+(y')^2}\,dx$
Derivative of $\arcsin x$	$\dfrac{1}{\sqrt{1-x^2}}$
Derivative of $\arctan x$	$\dfrac{1}{1+x^2}$
Improper integral test	$\displaystyle\int_a^\infty f(x)\,dx = \lim_{b\to\infty}\int_a^b f(x)\,dx$

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Using integration by parts, evaluate $\displaystyle\int x^3 e^{-x}\,dx$.

Solution

Let u = x^3$$dv = e^{-x}\,dx. du = 3x^2\,dx$$v = -e^{-x}.

$\int x^3 e^{-x}\,dx = -x^3 e^{-x} + 3\int x^2 e^{-x}\,dx$.

Repeating: $\int x^2 e^{-x}\,dx = -x^2 e^{-x} + 2\int xe^{-x}\,dx = -x^2 e^{-x} - 2xe^{-x} + 2\int e^{-x}\,dx$.

$= -x^2 e^{-x} - 2xe^{-x} - 2e^{-x}$.

Therefore: $\int x^3 e^{-x}\,dx = -x^3 e^{-x} - 3x^2 e^{-x} - 6xe^{-x} - 6e^{-x} + C$.

$\boxed{= -e^{-x}(x^3 + 3x^2 + 6x + 6) + C}$

Find the arc length of the curve $y = \ln(\cos x)$ from $x = 0$ to $x = \pi/4$.

Solution

$y' = -\tan x$. $1 + (y')^2 = 1 + \tan^2 x = \sec^2 x$.

$s = \displaystyle\int_0^{\pi/4} \sec x\,dx = [\ln|\sec x + \tan x|]_0^{\pi/4} = \ln(\sqrt{2}+1) - \ln(1) = \boxed{\ln(\sqrt{2}+1)}$.

Prove that $\displaystyle\int_0^{\pi/2} \sin^n x\,dx = \dfrac{n-1}{n} \cdot \dfrac{n-3}{n-2} \cdots \times \begin{cases} 1 & n \text{ odd} \\ \dfrac{\pi}{2} & n \text{ even}\end{cases}$ (Wallis’ formula).

Solution

Let $I_n = \displaystyle\int_0^{\pi/2} \sin^n x\,dx$.

Integration by parts with u = \sin^{n-1}x$$dv = \sin x\,dx:

$I_n = [-\cos x \sin^{n-1}x]_0^{\pi/2} + (n-1)\displaystyle\int_0^{\pi/2} \cos^2 x \sin^{n-2}x\,dx$

$= 0 + (n-1)\displaystyle\int_0^{\pi/2} (1-\sin^2 x)\sin^{n-2}x\,dx = (n-1)(I_{n-2} - I_n)$.

$nI_n = (n-1)I_{n-2}$ So $\boxed{I_n = \dfrac{n-1}{n}I_{n-2}}$.

Base cases: I_0 = \pi/2$$I_1 = 1.

For even $n$: $I_n = \dfrac{n-1}{n} \cdot \dfrac{n-3}{n-2} \cdots \dfrac{1}{2} \cdot \dfrac{\pi}{2}$.

For odd $n$: $I_n = \dfrac{n-1}{n} \cdot \dfrac{n-3}{n-2} \cdots \dfrac{2}{3} \cdot 1$. $\blacksquare$

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The gamma function extends the factorial: $\Gamma(n) = (n-1)!$ for positive integers, and $\Gamma(x) = \displaystyle\int_0^{\infty} t^{x-1}e^{-t}\,dt$ for $x > 0$.

Wallis’ formula leads to the important result: $\Gamma(1/2) = \sqrt{\pi}$.

For suitable functions $f$: $\displaystyle\int_0^{\infty} \frac{f(ax)-f(bx)}{x}\,dx = (f(0)-f(\infty))\ln\frac{b}{a}$.

Example: $\displaystyle\int_0^{\infty} \frac{e^{-ax}-e^{-bx}}{x}\,dx = \ln\frac{b}{a}$.

Leibniz’s rule: $\dfrac{d}{d\alpha}\displaystyle\int_a^b f(x,\alpha)\,dx = \int_a^b \frac{\partial f}{\partial\alpha}\,dx$.

This is a powerful technique for evaluating integrals that depend on a parameter.

If $0 \leq f(x) \leq g(x)$ for $x \geq a$ and $\displaystyle\int_a^{\infty} g(x)\,dx$ converges, Then $\displaystyle\int_a^{\infty} f(x)\,dx$ also converges.

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Evaluate $\displaystyle\int_0^{\infty} xe^{-x}\,dx$ and relate it to the mean of the exponential Distribution.

Solution

Integration by parts with u = x$$dv = e^{-x}\,dx:

$= [-xe^{-x}]_0^{\infty} + \displaystyle\int_0^{\infty} e^{-x}\,dx = 0 + 1 = \boxed{1}$.

This equals $E(X)$ for $X \sim \mathrm{Exp}(1)$ Confirming the result $E(X) = 1/\lambda$ with $\lambda = 1$.

Prove that $\displaystyle\int_0^{\pi/2} \sin^2 x\cos^2 x\,dx = \frac{\pi}{16}$.

Solution

$\sin^2 x\cos^2 x = \dfrac{\sin^2 2x}{4} = \dfrac{1-\cos 4x}{8}$.

$\displaystyle\int_0^{\pi/2} \frac{1-\cos 4x}{8}\,dx = \frac{1}{8}\!\left[x-\frac{\sin 4x}{4}\right]_0^{\pi/2} = \frac{1}{8}\cdot\frac{\pi}{2} = \boxed{\dfrac{\pi}{16}}$. $\blacksquare$

Use integration by parts twice to evaluate $\displaystyle\int e^x\cos x\,dx$.

Solution

$I = \displaystyle\int e^x\cos x\,dx = e^x\sin x - \int e^x\sin x\,dx = e^x\sin x - (-e^x\cos x + \int e^x\cos x\,dx)$.

$I = e^x\sin x + e^x\cos x - I$.

$2I = e^x(\sin x+\cos x)$.

$\boxed{I = \dfrac{e^x(\sin x+\cos x)}{2} + C}$

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Evaluate $\displaystyle\int_0^1 \frac{x^3}{1+x^2}\,dx$.

Solution

Let u = 1+x^2$$du = 2x\,dx. Note $x^2 = u-1$ So $x^3\,dx = x^2 \cdot x\,dx = (u-1)\cdot\dfrac{du}{2}$.

$\displaystyle\int_0^1 \frac{x^3}{1+x^2}\,dx = \frac{1}{2}\int_1^2 \frac{u-1}{u}\,du = \frac{1}{2}\int_1^2 \!\left(1-\frac{1}{u}\right)du$

$= \frac{1}{2}\Big[u-\ln u\Big]_1^2 = \frac{1}{2}(2-\ln 2 - 1) = \boxed{\frac{1}{2}(1-\ln 2)}$.

Prove that the function $F(x) = \displaystyle\int_0^x \frac{dt}{1+t^4}$ is increasing and Bounded above.

Solution

$F'(x) = \dfrac{1}{1+x^4} > 0$ for all $x \geq 0$ So $F$ is strictly increasing. ✓

$F(x) < F(\infty) = \displaystyle\int_0^{\infty} \frac{dt}{1+t^4} < \int_0^{\infty} \frac{dt}{1+t^2} = \frac{\pi}{2}$. ✓

Therefore $F$ is increasing and bounded above by $\pi/2$. $\blacksquare$

• Pure Mathematics: Further calculus is a core pure topic
• Hyperbolic Functions: Integration covers hyperbolic functions
• Differential Equations: Calculus solves differential equations