Methods in Calculus (Edexcel A Level Further Maths: Further Pure 1): Exam Questions

Use L'Hôpital's rule to show that

$\underset{x\to 0}{\text{lim}}\left(\frac{1}{\text{sin}x}-\frac{1}{x}\right)=0$

Use the substitution $t=\text{tan}\frac{\theta }{2}$ to determine the exact value of

${\int }_0^{\frac{\pi }{2}}\frac{5}{4+2\text{cos}\theta }\text{d}\theta$

giving your answer in simplest form.

$y={\text{e}}^{3x}\text{sin}x$

Use Leibnitz's theorem to show that

$\frac{{\text{d}}^{4}y}{\text{d}{x}^4}=28{\text{e}}^{3x}\text{sin}x+96{\text{e}}^{3x}\text{cos}x$

Hence express $\frac{{\text{d}}^{4}y}{\text{d}{x}^4}$ in the form

$R{\text{e}}^{3x}\text{sin }\left(x+\alpha \right)$

where $R$ and $\alpha$ are constants to be determined, $R>0$ and $0<\alpha <\frac{\pi }{2}$

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

Use the substitution $t=\text{tan}\frac{\theta }{2}$ to show that

$\int \frac{1}{2\text{sin}\theta +\text{cos}\theta +2}\text{d}\theta =\int \frac{a}{t^2+bt+c}\text{d}t$

where $a$, $b$ and $c$ are constants to be determined.

Hence show that

${\int }_{\frac{\pi }{2}}^{\frac{2\pi }{3}}\frac{1}{2\text{sin}\theta +\text{cos}\theta +2}\text{d}\theta =\text{ln}\left(\frac{2\sqrt{3}}{3}\right)$

Show that the substitution $t=\text{tan}\left(\frac{x}{2}\right)$ transforms the integral

$\int \frac{1}{2\text{sin}x-\text{cos}x+5}\text{d}x$

into the integral

$\int \frac{1}{3t^2+2t+2}\text{d}t$

Hence determine

$\int \frac{1}{2\text{sin}x-\text{cos}x+5}\text{ d}x$

Use L'Hospital's rule to determine

$\underset{x\to 0}{\text{lim}}\left(\frac{1}{(x+3)\text{tan}6x·\text{cosec}2x}\right)$

(Solutions relying entirely on calculator technology are not acceptable.)

Use l'Hospital's Rule to show that

$\underset{x\to \frac{\pi }{2}}{\lim }\frac{\left({\text{e}}^{\text{sin}x}-\text{cos}\left(3x\right)-\text{e}\right)}{\text{tan}\left(2x\right)}=-\frac{3}{2}$

$\text{f}\left(x\right)=x^4\text{sin}\left(2x\right)$

Use Leibnitz's theorem to show that the coefficient of $(x-\pi {)}^8$ in the Taylor series expansion of $\text{f}\left(x\right)$ about $\pi$ is

$\frac{a\pi +b{\pi }^3}{315}$

where $a$ and $b$ are integers to be determined.

The Taylor series expansion of $\text{f}\left(x\right)$ about $x=k$ is given by

$\text{f}\left(x\right)=\text{f}\left(k\right)+(x-k)\text{f}\text{'}\left(k\right)+\frac{(x-k{)}^2}{2!}\text{f}\text{'}\text{'}\left(k\right)+\dots +\frac{(x-k{)}^r}{r!}{\text{f}}^{\left(r\right)}\left(k\right)+\dots$

Given that $k$ is a real non-zero constant and that

$y=x^3\text{sin}kx$

use Leibnitz's theorem to show that

$\frac{{\text{d}}^{5}y}{\text{d}{x}^5}=(k^2{x}^2+A)k^{3}x\text{cos}kx+B(k^2{x}^2+C)k^2\text{sin}kx$

where $A$, $B$ and $C$ are integers to be determined.

$I=\int \frac{1}{4\text{cos}x-3\text{sin}x}\text{d}x0<x<\frac{\pi }{4}$

Use the substitution $t=\text{tan}\left(\frac{x}{2}\right)$ to show that

$I=\frac{1}{5}\text{ln}\left(\frac{2+\text{tan}\left(\frac{x}{2}\right)}{1-2\text{tan}\left(\frac{x}{2}\right)}\right)+k$

where $k$ is an arbitrary constant.

Given $k$ is a constant and that

$y=x^3{\text{e}}^{kx}$

use Leibnitz theorem to show that

$\frac{{\text{d}}^{n}y}{\text{d}{x}^n}=k^{n-3}{\text{e}}^{kx}\left(k^3{x}^3+3n{k}^2{x}^2+3n(n-1)kx+n(n-1)(n-2)\right)$

Use L'Hospital's rule to show that

$\underset{x\to 0}{\lim }\left(\frac{1}{x}-\frac{1}{{\text{e}}^x-1}\right)=\frac{1}{2}$

$y={\text{e}}^{2x}\text{sin}x$

Use Leibnitz's theorem to show that

$\frac{{\text{d}}^{4}y}{\text{d}{x}^4}={\text{e}}^{2x}(24\text{cos}x-7\text{sin}x)$

Hence express $\frac{{\text{d}}^{4}y}{\text{d}{x}^4}$ in the form $R{\text{e}}^{2x}\text{cos}(x+\alpha )$, where $R$ and $\alpha$ are constants to be determined, $R>0$ and $0<\alpha <\frac{\pi }{2}$.

In this question you must show all stages of your working.

Solutions relying on calculator technology are not acceptable.

Use the substitution $t=\text{tan}\frac{\theta }{2}$ to show that

$\int \frac{1}{2+\text{sin}\theta }\text{d}\theta =\int \frac{a}{t^2+bt+c}\text{d}t$

where $a$, $b$ and $c$ are constants to be determined.

Hence show that

${\int }_0^{\frac{\pi }{2}}\frac{1}{2+\text{sin}\theta }\text{d}\theta =\frac{\sqrt{3}\pi }{9}$