Question
The acceleration, a ms-2 , of a particle moving in a horizontal line at time t seconds, t ≥ 0 , is given by a = – (1+v) where v ms-1 is the particle’s velocity and v > -1.
At t = 0 , the particle is at a fixed origin O and has initial velocity v0 ms-1 .
(a) By solving an appropriate differential equation, show that the particle’s velocity at time t is given by v ( t ) = (1 + v0) e-t – 1 . [6]
(b) Initially at O, the particle moves in the positive direction until it reaches its maximum displacement from O. The particle then returns to O.
Let s metres represent the particle’s displacement from O and smax its maximum displacement from O.
(i) Show that the time T taken for the particle to reach smax satisfies the equation eT = 1 + v0 .
(ii) By solving an appropriate differential equation and using the result from part (b) (i), find an expression for smax in terms of v0 . [7]
Let v (T – k) represent the particle’s velocity k seconds before it reaches smax , where v (T – k) = (1 + v0) e-(T – k) – 1 .
(c) By using the result to part (b) (i), show that v (T – k) = ek – 1 . [2]
Similarly, let v (T + k) represent the particle’s velocity k seconds after it reaches smax .
(d) Deduce a similar expression for v (T + k) in terms of k . [2]
(e) Hence, show that v (T – k) + v (T + k) ≥ 0 . [3]
▶️Answer/Explanation
Ans:
(a) Since $a=\frac{\text{d}v}{\text{d}t}$, and $a=-\left(1+v\right)$, we have
$$\begin{eqnarray}
\frac{\text{d}v}{\text{d}t} = -1-v \nonumber \\
\frac{\text{d}v}{\text{d}t}+v = -1.
\end{eqnarray}$$
Let $\text{I}\left(x\right)=\text{e}^{\int 1 \text{d}t}=\text{e}^t$.<br>
Then, we have
$$\begin{eqnarray}
\frac{\text{d}}{\text{d}t}\left(v\text{e}^t\right) = -\text{e}^t \nonumber \\
v\text{e}^t = -\int\text{e}^t \text{d}t \nonumber \\
v\text{e}^t = -\text{e}^t+c \nonumber \\
v = c\text{e}^{-t}-1.
\end{eqnarray}$$
Since $v=v_0$ when $t=0$, we have $c=1+v_0$, i.e., $v=\left(1+v_0\right)\text{e}^{-t}-1$.<br>
(b)(i) At $s_{\text{max}}$, $t=T$, i.e., we have
$$\begin{eqnarray}
v\left(T\right) = 0 \nonumber \\
\left(1+v_0\right)\text{e}^{-T}-1 = 0 \nonumber \\
\left(1+v_0\right)\text{e}^{-T} = 1 \nonumber \\
1+v_0 = \text{e}^T.
\end{eqnarray}$$
(b)(ii) Since $\frac{\text{d}s}{\text{d}t}=\left(1+v_0\right)\text{e}^{-t}-1$, integrating both sides with respect to $t$, we have
$$\begin{eqnarray}
s &=& \int\left(1+v_0\right)\text{e}^{-t}-1\text{d}t \nonumber \\
&=& -\left(1+v_0\right)\text{e}^{-t}-t+c.
\end{eqnarray}$$
When $t=0$, $s=0$, i.e.,
$$\begin{eqnarray}
0 = -\left(1+v_0\right)+c \nonumber \\
c = \left(1+v_0\right).
\end{eqnarray}$$
Thus, $s=\left(1+v_0\right)-\left(1+v_0\right)\text{e}^{-t}-t$.<br>
Since from (b)(i) we have $1+v_0=\text{e}^T$, $T=\ln\left(1+v_0\right)$, i.e., $s_{\text{max}}=v_0-\ln \left(1+v_0\right)$.<br>
(c) Since $\text{e}^T=1+v_0$, we have $\left(1+v_0\right)\text{e}^{-T}=1$, i.e.,
$$\begin{eqnarray}
v\left(T-k\right) &=& \left(1+v_0\right)\text{e}^{-\left(T-k\right)}-1 \nonumber \\
&=& \left(1+v_0\right)\text{e}^{-T}\text{e}^k-1 \nonumber \\
&=& \text{e}^k-1.
\end{eqnarray}$$
(d) Similarly, as in (c),
$$\begin{eqnarray}
v\left(T+k\right) &=& \left(1+v_0\right)\text{e}^{-\left(T+k\right)}-1 \nonumber \\
&=& \left(1+v_0\right)\text{e}^{-T}\text{e}^{-k}-1 \nonumber \\
&=& \text{e}^{-k}-1.
\end{eqnarray}$$
(e)
$$\begin{eqnarray}
v\left(T-k\right)+v\left(T+k\right) &=& \text{e}^k-1+\text{e}^{-k}-1 \nonumber \\
&=& \text{e}^k+\text{e}^{-k}-2 \nonumber \\
&=& \frac{\text{e}^{2k}-2\text{e}^k+1}{\text{e}^k} \nonumber \\
&=& \frac{\left(\text{e}^k-1\right)^2}{\text{e}^k} \geq 0.
\end{eqnarray}$$
Question
Find \(\int {(1 + {{\tan }^2}x){\text{d}}x} \).[2]
Find \(\int {{{\sin }^2}x{\text{d}}x} \).[3]
▶️Answer/Explanation
Markscheme
\(\int {(1 + {{\tan }^2}x){\text{d}}x} = \int {{{\sec }^2}x{\text{d}}x = \tan x( + c)} \) M1A1
[2 marks]
\(\int {{{\sin }^2}x{\text{d}}x} = \int {\frac{{1 – \cos 2x}}{2}{\text{d}}x} \) M1A1
\( = \frac{x}{2} – \frac{{\sin 2x}}{4}( + c)\) A1
Note: Allow integration by parts followed by trig identity.
Award M1 for parts, A1 for trig identity, A1 final answer.
[3 marks]
Total [5 marks]
Question
A function \(f\) is defined by \(f(x) = \frac{{3x – 2}}{{2x – 1}},{\text{ }}x \in \mathbb{R},{\text{ }}x \ne \frac{1}{2}\).
Find an expression for \({f^{ – 1}}(x)\).[4]
Given that \(f(x)\) can be written in the form \(f(x) = A + \frac{B}{{2x – 1}}\), find the values of the constants \(A\) and \(B\).[2]
Hence, write down \(\int {\frac{{3x – 2}}{{2x – 1}}} {\text{d}}x\).[1]
▶️Answer/Explanation
Markscheme
\(f:x \to y = \frac{{3x – 2}}{{2x – 1}}\;\;\;{f^{ – 1}}:y \to x\)
\(y = \frac{{3x – 2}}{{2x – 1}} \Rightarrow 3x – 2 = 2xy – y\) M1
\( \Rightarrow 3x – 2xy = – y + 2\) M1
\(x(3 – 2y) = 2 – y\)
\(x = \frac{{2 – y}}{{3 – 2y}}\) A1
\(\left( {{f^{ – 1}}(y) = \frac{{2 – y}}{{3 – 2y}}} \right)\)
\({f^{ – 1}}(x) = \frac{{2 – x}}{{3 – 2x}}\;\;\;\left( {x \ne \frac{3}{2}} \right)\) A1
Note: \(x\) and \(y\) might be interchanged earlier.
Note: First M1 is for interchange of variables second M1 for manipulation
Note: Final answer must be a function of \(x\)
[4 marks]
\(\frac{{3x – 2}}{{2x – 1}} = A + \frac{B}{{2x – 1}} \Rightarrow 3x – 2 = A(2x – 1) + B\)
equating coefficients \(3 = 2A\) and \( – 2 = – A + B\) (M1)
\(A = \frac{3}{2}\) and \(B = – \frac{1}{2}\) A1
Note: Could also be done by division or substitution of values.
[2 marks]
\(\int {f(x){\text{d}}x = \frac{3}{2}x – \frac{1}{4}\ln \left| {2x – 1} \right| + c} \) A1
Note: accept equivalent e.g. \(\ln \left| {4x – 2} \right|\)
[1 mark]
Total [7 marks]
Question
Find
(a) \(\int (2x+1)^{3}\: dx\) (b) \(\int (2x^{2}+1)^{3}\: dx\) (c) \(\int \frac{(2x^{2}+1)^{3}}{x}dx\)
▶️Answer/Explanation
Ans
(a) \(\int (2x+1)^{3}dx=\frac{(2x+1)^{4}}{8}+c\)
(b) \(\int (2x^{2}+1)^{3}\, dx=\int (8x^{6}+12x^{4}+6x^{2}+1)dx=\frac{8x^{7}}{7}+\frac{12x^{5}}{5}+2x^{3}+x+c\)
(c) \(\int \frac{(2x^{2}+1)^{3}}{x}dx=\int (8x^{5}+12x^{3}+6x+\frac{1}{x})dx=\frac{8x^{6}}{3}+3x^{4}+3x^{2}+\textup{In}\left | x \right |+c\)
Question
Use the double angle identities to calculate
(a) \(\int \textup{cos}^{2}\; xdx\) (b) \(\int \textup{sin}^{2}\; xdx\) (c)\(\int \textup{sin}\, x\, \textup{cos}\, xdx\) (d) \(\int \textup{sin}\, 5x\, \textup{cos}\, 5xdx\)
▶️Answer/Explanation
Ans
(a) \(\int \textup{cos}^{2}\: xdx=\frac{1}{2}\int (1+\textup{cos2}x)dx=\frac{x}{2}+\frac{\textup{sin2}x}{4}+c\)
(b) \(\int \textup{sin}^{2}\: xdx=\frac{1}{2}\int (1-\textup{cos2}x)dx=\frac{x}{2}-\frac{\textup{sin2}x}{4}+c\)
(c) \(\int \textup{sin}\, x\, \textup{cos}\, xdx=\frac{1}{2}\int \textup{sin}\, 2xdx=-\frac{\textup{cos}2x}{4}+c\)
(d) \(\int \textup{sin}\, 5x\, \textup{cos}\, 5xdx=\frac{1}{2}\int \textup{sin}10xdx=-\frac{\textup{cos}10x}{20}+c\)
Questions
Find
(a) \(\int \sqrt{x}(x^{3}-1)dx\) (b) \(\int \frac{3x^{2}-x+2\sqrt{x}}{3\sqrt{x}}dx\)
▶️Answer/Explanation
Ans
(a) \(\int \sqrt{x}(x^{3}-1)dx=\int (x^{3.5}-x^{0.5})dx=\frac{x^{4.5}}{4.5}-\frac{x^{1.5}}{1.5}+c\)
(b) \(\int \frac{3x^{2}-x+2\sqrt{2}}{3\sqrt{x}}dx=\int (x^{1.5}-\frac{1}{3}x^{0.5}+\frac{2}{3})dx=\frac{x^{2.5}}{2.5}-\frac{x^{1.5}}{4.5}+\frac{2}{3}x+c\)
Question
Find (a) \(\int \frac{e^{3x}+2e^{x}+4}{3e^{2x}}dx\) (b) \(\int \frac{8^{x}+2^{x}+1}{4^{x}}dx\).
▶️Answer/Explanation
Ans
(a) \(\int \frac{e^{3x}+2e^{x}+4}{3e^{2x}}dx=\int (\frac{1}{3}e^{x}-\frac{2}{3}e^{-x}+\frac{4}{3}e^{-2x})dx=\frac{1}{3}e^{x}+\frac{2}{3}e^{-x}-\frac{2}{3}e^{-2x}+c\)
(b) \(\int \frac{8^{x}+2^{x}+1}{4^{x}}dx=\int (2^{x}-2^{-x}+4^{-x})dx=\frac{2^{x}}{\textup{In}\, 2}+\frac{2^{-x}}{\textup{In}\, 2}-\frac{4^{-x}}{\textup{In}\, 4}+c\)
Question
Let \({f}'(x)=2e^{-x}+10\, \textup{sin}\, 5x+1\).
Find \(f (x)\) , given that the curve of this function passes through the point A(0,5).
▶️Answer/Explanation
Ans
\(f(x)=-2e^{-x}-2\textup{cos}5x+x+c\)
\(f(0)=5\Leftrightarrow -2-2+0+c=5\Leftrightarrow c=9\)
Hence, \(f(x)=-2e^{-x}-2\textup{cos}5x+x+9\)
Question
Let \({f}”(x)=6ax+b\).
Find \(f (x)\) , given that the function has a maximum at A(1,5) and a point of inflection at B(0,3)
▶️Answer/Explanation
Ans
\({f}”(x)=6ax+b\), \({f}'(x)=3ax^{2}+bx+c\) and \(f(x)=ax^{3}+\frac{b}{2}x^{2}+cx+d\)
Therefore, \(f(x)=-x^{3}+3x+3\)
Question
Find
(a) \(\int \frac{7}{16+x^{2}}dx\) (b) \(\int \frac{7}{1+16x^{2}}dx\) (c) \(\int \frac{7}{25+16x^{2}}dx\) (d) \(\int \frac{7}{2+x^{2}}dx\) (e) \(\int \frac{7}{2+3x^{2}}dx\)
▶️Answer/Explanation
Question
Find
(a) \(\int \frac{7}{\sqrt{16-x^{2}}}dx\) (b) \(\int \frac{7}{\sqrt{1-16x^{2}}}dx\) (c) \(\int \frac{7}{\sqrt{25-16x^{2}}}dx\) (d) \(\int \frac{7}{\sqrt{2-x^{2}}}dx\) (e) \(\int \frac{7}{\sqrt{2-3x^{2}}}dx\)
▶️Answer/Explanation
Question
Find (a) \(\int \frac{7}{(x-1)^{2}+4}dx\) (b) \(\int \frac{7}{4(x-1)^{2}+1}dx\)
▶️Answer/Explanation
Ans
(a) \(\int \frac{7}{(x-1)^{2}+4}dx=\frac{7}{2}\textup{arctan}\frac{x-1}{2}+c\) (b) \(\int \frac{7}{4(x-1)^{2}+1}dx=\frac{7}{2}\textup{arctan}(2x-2)+c\)
Question
Find (a) \(\int \frac{7}{\sqrt{4-(x-1)^{2}}}dx\) (b) \(\int \frac{7}{\sqrt{1-4(x-1)^{2}}}dx\)
▶️Answer/Explanation
Ans
(a) \(\int \frac{7}{\sqrt{4-(x-1)^{2}}}dx=7\textup{arcsin}\frac{x-1}{2}+c\) (b) \(\int \frac{7}{\sqrt{1-4(x-1)^{2}}}dx=\frac{7}{2}\textup{arcsin}(2x-2)+c\)
Question
(a) Express \(\frac{2}{x^{2}-10x+24}\) in partial fractions
(b) Express \(x^{2}-10x+26\) in vertex form
(c) Find the integrals
(i) \(\int \frac{2}{x^{2}-10x+24}dx\) (ii) \(\int \frac{2}{x^{2}-10x+25}dx\) (iii) \(\int \frac{2}{x^{2}-10x+26}dx\)
▶️Answer/Explanation
Ans
(a) \(\frac{2}{x^{2}-10x+24}=\frac{A}{x-6}+\frac{B}{x-4}\Rightarrow 2=A(x-4)+B(x-6)\)
For \(x=6,\, 2A=2\Leftrightarrow A=1\)
For \(x=4,\, -2B=2\Leftrightarrow B=-1\)
Hence, \(\frac{2}{x^{2}-10x+24}=\frac{1}{x-6}-\frac{1}{x-4}\)
(b) \(x^{2}-10x+26=(x-5)^{2}+1\)
(c) (i) \(\int \frac{2}{x^{2}-10x+24}dx=\int \frac{1}{x-6}-\frac{1}{x-4}dx=\textup{In}\left | x-6 \right |-\textup{In}\left | x-4 \right |+c=\, \textup{In}\left | \frac{x-6}{x-4} \right |+c\)
(ii) \(\int \frac{2}{x^{2}-10x+25}dx=\int \frac{2}{(x-5)^{2}}dx=-\frac{2}{x-5}+c\)
(iii) \(\int \frac{2}{x^{2}-10x+26}dx=\int \frac{2}{(x-5)^{2}+1}dx=2\, \textup{arctan}(x-5)+c\)
Question
Let \(f(x)=\frac{x^{2}+3x+12}{x(x+2)^{2}}\)
(a) Show that \(f(x)=\frac{3}{x}-\frac{2}{(x+2)}-\frac{5}{(x+2)^{2}}\)
(b) Hence find \(\int f(x)dx\)
▶️Answer/Explanation
Ans
(b) \(\int f(x)dx=3\, \textup{In}\left | x \right |-2\, \textup{In}\left | x+2 \right |+\frac{5}{x+2}+c\)
Question
Complete the following table. [Look at the formula booklet!]
| Integral | Result |
| \(\int \textup{sec}^{2}\, xdx\) | |
| \(\int (\textup{tan}^{2}\, x+1)dx\) | |
| \(\int \frac{5}{\textup{cos}^{2}\, x}dx\) | |
| \(\int \textup{csc}^{2}\, xdx\) | |
| \(\int (\textup{cot}^{2}\, x+1)dx\) | |
| \(\int \frac{5}{\textup{sin}^{2}\, x}dx\) | |
| \(\int \textup{sec}\, x\, \textup{tan}\, xdx\) | |
| \(\int \textup{csc}\, x\, \textup{cot}\, xdx\) | |
| \(\int \textup{tan}^{2}\, xdx\) | |
| \(\int \textup{cot}^{2}\, xdx\) |
▶️Answer/Explanation
Ans
| Integral | Result |
| \(\int \textup{sec}^{2}\, xdx\) | \(\textup{tan}\, x+c\) |
| \(\int (\textup{tan}^{2}\, x+1)dx\) | \(\textup{tan}\, x+c\) |
| \(\int \frac{5}{\textup{cos}^{2}\, x}dx\) | \(5\, \textup{tan}\, x+c\) |
| \(\int \textup{csc}^{2}\, xdx\) | \(-\textup{cot}\, x+c\) |
| \(\int (\textup{cot}^{2}\, x+1)dx\) | \(-\textup{cot}\, x+c\) |
| Integral | Result |
| \(\int \frac{5}{\textup{sin}^{2}\, x}dx\) | \(-5\, \textup{cot}\, x+c\) |
| \(\int \textup{sec}\, x\, \textup{tan}\, xdx\) | \(\textup{sec}\, x+c\) |
| \(\int \textup{csc}\, x\, \textup{cot}\, xdx\) | \(-\, \textup{csc}\, x+c\) |
| \(\int \textup{tan}^{2}\, xdx\) | \(\textup{tan}\, x\, -x+c\) |
| \(\int \textup{cot}^{2}\, xdx\) | \(-\textup{cot}\, x\, -x+c\) |
Question
Let \(f(t)=t^{\frac{1}{3}}\left ( 1-\frac{1}{2t^{\frac{5}{3}}} \right )\). Find \(\int f(t)\, \textup{d}t\). (Total 3 marks)
▶️Answer/Explanation
Ans
\(\int t^{\frac{1}{3}}\left ( 1-\frac{1}{2t^{\frac{2}{5}}} \right )\textup{d}t=\int t^{\frac{1}{3}}\left ( 1-\frac{t^{\frac{5}{3}}}{2} \right )\textup{d}t=\int \left ( t^{\frac{1}{3}}-\frac{t^{\frac{4}{3}}}{2} \right )\textup{d}t\) (M1)
\(=\frac{3}{4}t^{\frac{4}{3}}+\frac{3}{2}t^{\frac{1}{3}}+c\) (M1)(A1) (C3)
Note: Do not penalise for the absence of +C. [3]
Question
The function \({f}’\) is given by \({f}'(x)=2\, \textup{sin}\left ( 5x-\frac{\pi }{2} \right )\).
(a) Write down \({f}”(x)\). (b) Given that \(f\left ( \frac{\pi}{2} \right )=1\), find \(f(x)\).
(Total 6 marks)
▶️Answer/Explanation
Ans
(a) Using the chain rule \({f}'(x)=\left ( 2\, \textup{cos}\left ( 5x-\frac{\pi }{2} \right ) \right )5\) (M1)
\(=10\, \textup{cos}\left ( 5x-\frac{\pi }{2} \right )\) A1 2
(b) \(f(x)=\int {f}'(x)dx=-\frac{2}{5}\textup{cos}\left ( 5x-\frac{\pi }{2} \right )+c\) A1
Substituting to find \(c,f\left ( \frac{\pi }{2} \right )=\frac{2}{5}\textup{cos}\left ( 5\left ( \frac{\pi }{2} \right )-\frac{\pi }{2} \right )+c=1\) M1
\(c=1+\frac{2}{5}\textup{cos}2\pi =1+\frac{2}{5}=\frac{7}{5}\) (A1)
\(f(x)=\frac{2}{5}\textup{cos}\left ( 5x-\frac{\pi }{2} \right )+\frac{7}{5}\) A1 4 [6]
Question
Find the indefinite integrals
(a) \(\int \frac{1}{x^{2}+2}dx\) (b) \(\int \frac{x}{x^{2}+2}dx\) (c) \(\int \frac{x^{2}}{x^{2}+2}dx\) (d) \(\int \frac{x^{3}}{x^{2}+2}dx\)
▶️Answer/Explanation
Ans
(a) \(\frac{1}{\sqrt{2}}\textup{arctan}\frac{x}{\sqrt{2}}+c\) (b) \(\frac{1}{2}\textup{In}(x^{2}+2)+c\) (c) \(x-\frac{2}{\sqrt{2}}\textup{arctan}\frac{x}{\sqrt{2}}+c\) (d) \(\frac{x^{2}}{2}-\textup{In}(x^{2}+2)+c\)
Question
Find the indefinite integral \(\int \frac{12x+10}{3x^{2}+5x+1}dx\)
▶️Answer/Explanation
Ans
\(2\, \textup{In}(3x^{2}+5x+1)+c\)
Question
Find the indefinite integrals (a) \(\int \textup{tan}\, xdx\) (b) \(\int \textup{cot}\, xdx\)
▶️Answer/Explanation
Ans
(a) \(\int \textup{tan}\, xdx=\int \frac{\textup{sin}\, x}{\textup{cos}\, x}dx=-\textup{In}\left | \textup{cos}\, x \right |+c\) (use the substitution \(u=\textup{cos}\, x\))
(b) \(\int \textup{cot}\, xdx=\int \frac{\textup{cos}\, x}{\textup{sin}\, x}dx=\textup{In}\left | \textup{sin}\, x \right |+c\) (use the substitution \(u=\textup{sin}\, x\))
Question
Find the following indefinite integrals by using the substitution \(u=x^{2}+5\)
(a) \(\int \frac{x^{5}}{x^{2}+5}dx\) (b) \(\int \frac{x^{3}}{\sqrt{x^{2}+5}}dx\) (c) \(\int x^{3}(x^{2}+5)^{5}dx\)
▶️Answer/Explanation
Ans
(a) \(\frac{25}{2}\textup{In}(x^{2}+5)+\frac{(x^{2}+5)^{2}}{4}-5x^{2}+c\) (b) \(\frac{(x^{2}-10)\sqrt{x^{2}+5}}{3}+c\) (c) \(\frac{(x^{2}+5)^{6}(6x^{2}-5)}{84}+c\)
Question
Find the indefinite integrals
(a) \(\int \frac{x}{x^{2}+9}dx\) (b) \(\int \frac{x}{(x^{2}+9)^{2}}dx\) (c) \(\int \frac{x^{2}+x+12}{x^{2}+9}dx\)
▶️Answer/Explanation
Ans
(a) \(\frac{1}{2}\textup{In}(x^{2}+9)+c\) (b) \(-\frac{1}{2(x^{2}+9)}+c\) (c) \(\frac{1}{2}\textup{In}(x^{2}+9)+x+\textup{arctan}\frac{x}{3}+c\)
Question
Find the indefinite integrals
(a) \(\int \textup{tan}^{5}\, x\, \textup{sec}^{2}\, xdx\) (b) \(\int \frac{\textup{sec}^{2}\, x}{\sqrt{\textup{tan}\, x}}dx\) (c) \(\int \frac{\textup{sec}^{2}\, x}{e^{\textup{tan}\, x}}dx\) (d) \(\int \frac{\textup{sin}^{5}\, x}{\textup{cos}^{7}\, x}dx\)
▶️Answer/Explanation
Ans
(a) \(\frac{\textup{tan}^{6}\, x}{6}+c\) (b) \(2\sqrt{\textup{tan}\, x}+c\) (c) \(-\frac{1}{e^{\textup{tan}\, x}}+c\) (d) \(\frac{\textup{tan}^{6}\, x}{6}+c\)
Question
(a) Calculate \(\int \frac{1}{5+x^{2}}dx\) , by using the substitution \(x=\sqrt{5}\, \textup{tan}\, \theta \)
(b) Calculate (i) \(\int \frac{1}{\sqrt{5-x^{2}}}dx\) , (ii) \(\int \sqrt{5-x^{2}\, dx}\) by using the substitution \(x=\sqrt{5}\, \textup{sin}\, \theta \)
▶️Answer/Explanation
Ans
(a) \(\frac{1}{\sqrt{5}}\textup{arctan}\frac{x}{\sqrt{5}}+c\) (b) (i) \(\textup{arcsin}\frac{x}{\sqrt{5}}+c\) , (ii) \(\frac{5}{2}\textup{arcsin}\frac{x}{\sqrt{5}}+\frac{x\sqrt{5-x^{2}}}{2}+c\)
Question
Calculate \(\int \sqrt{\frac{x}{4-x}}dx\) , by using the substitution \(x=4\, \textup{sin}^{2}\, \theta \)
Express your answer in the form \(A\, \textup{arcsin}\frac{\sqrt{x}}{2}+B\sqrt{4x-x^{2}}+c\)
▶️Answer/Explanation
Ans
\(4\, \textup{arcsin}\frac{\sqrt{x}}{2}-\sqrt{4x-x^{2}}+c\)
Question
(a) Show that \(\frac{2x+4}{(x^{2}+4)(x-2)}=\frac{1}{x-2}-\frac{x}{x^{2}+4}\)
(b) Hence find \(\int \frac{2x+4}{(x^{2}+4)(x-2)}dx\)
▶️Answer/Explanation
Ans
(b) \(\textup{In}(x-2)-\frac{1}{2}\textup{In}(x^{2}+4)+c\)
Question
Using the substitution \(y=2-x\), or otherwise, find \(\int \left ( \frac{x}{2-x} \right )^{2}dx\). (Total 6 marks)
▶️Answer/Explanation
Ans
\(I=\int \left ( \frac{2-y}{y} \right )^{2}(-dy)\) (M1)(A1)
\(=\int \left ( \frac{4}{y^{2}}-\frac{4}{y}+1 \right )\textup{d}y=\frac{4}{y}+4\, \textup{In}\left | y \right |-y+c\) (A1)(A1)(A1)
\(=\frac{4}{2-x}+4\textup{In}\left | 2-x \right |-(2-x)+c\) (A1) (C6)
Note: c and modulus signs not required. [6]
Question
Using the substitution \(u=\frac{1}{2}x+1\),or otherwise, find the integral \(\int x\sqrt{\frac{1}{2}x+1}\: dx\). (Total 4 marks)
▶️Answer/Explanation
Ans
Let \(u=\frac{1}{2}x+1\Leftrightarrow x=2(u-1)\Rightarrow \frac{\textup{d}x}{\textup{d}u}=2\)
Then \(\int x\left ( \frac{1}{2}x+1 \right )^{1/2}dx\)
\(=\int 2(u-1)\times u^{1/2}\times 2\textup{d}u\) (M1)
\(=4\int (u^{3/2}-u^{1/2})\textup{d}u\) (A1)
\(=4\left [ \frac{2}{5}u^{5/2}-\frac{2}{3}u^{3/2} \right ]+C\) (M1)
\(=4\left [ \frac{2}{5}\left ( \frac{1}{2}x+1 \right )^{5/2}-\frac{2}{3}\left ( \frac{1}{2}x+1 \right )^{3/2} \right ]+C\) (A1) [4]
Question
Find \(\int \frac{e^{x}}{e^{2x}+9}dx\). (Total 5 marks)
Extra question
Find in a similar way the integral \(\int \frac{3^{x}}{9^{x}+9}dx\)
▶️Answer/Explanation
Ans
\(\textup{Let }u=e^{x}\) M1
\(\textup{d}u=e^{x}\textup{d}x\, (\textup{or equivalent})\) A1
\(\textup{When}\, x=0,u=1\, \textbf{and}\, \textup{when}\, x=\textup{In}\, 3, u=3\) (A1)
\(\int_{0}^{\textup{In}3}\frac{e^{x}}{e^{2x}+9}dx=\int_{1}^{3}\frac{1}{u^{2}+9}\textup{d}u\) A1
\(=\frac{1}{3}\left [ \textup{arctan}\frac{u}{3} \right ]^{3}_{1}\) A1
\(=\frac{1}{3}\left ( \textup{arctan}1-\textup{arctan}\frac{1}{3} \right )\left ( =\frac{\pi }{12}-\frac{1}{3}\textup{arctan}\frac{1}{3},\frac{1}{3}\textup{arctan}\frac{1}{2} \right )\) A1 N0 [6]
Extra question
Find in a similar way the integral \(\int \frac{3^{x}}{9^{x}+9}dx\)
Question
Use the substitution \(u=x+2\) to find \(\int \frac{x^{3}}{(x+2)^{2}}\textup{d}x\). (Total 6 marks)
▶️Answer/Explanation
Ans
Substituting \(u=x+2\Rightarrow u-2=x,\textup{d}u=\textup{d}x\) (M1)
\(\int \frac{x^{3}}{(x+2)^{2}}\textup{d}x=\frac{(u-2)^{3}}{u^{2}}\textup{d}u\) A1
\(=\int \frac{u^{3}-6u^{2}+12u-8}{u^{2}}\textup{d}u\) A1
\(\int u\, \textup{d}u+\int (-6)\textup{d}u+\int \frac{12}{u}\textup{d}u+\int 8u^{-2}\textup{d}u\) A1
\(=\frac{u^{2}}{2}-6u+12\textup{In}\left | u \right |+8u^{-1}+c\) A1
\(=\frac{(x+2)^{2}}{2}-6(x+2)+12\, \textup{In}\left | x+2 \right |+\frac{8}{x+2}+c\) A1 [6]
Question
By using an appropriate substitution find \(\int \frac{\textup{tan}(\textup{In}\, y)}{y}dy,y> 0\). (Total 6 marks)
Extra question
The integral \(\int \frac{\textup{cot}(\textup{In}\, y)}{y}dy,y> 0\) can be found in a similar way. Write down the result!
▶️Answer/Explanation
Ans
\(\textup{Let}\, u=\textup{In}\, y\Rightarrow \textup{d}u=\frac{1}{y}\textup{d}y\) A1(A1)
\(\int \frac{\textup{tan}(\textup{In}\, y)}{y}\textup{d}y=\int \textup{tan}\, u\, \textup{d}u\) A1
\(\int \frac{\textup{sin}\, u}{\textup{cos}\, u}\textup{d}u=-\textup{In}\left | \textup{cos}\, u \right |+c\) A1
EITHER
\(\int \frac{\textup{tan}(\textup{In}\, y)}{y}\textup{d}y=-\textup{In}\left | \textup{cos}(\textup{In}\, y) \right |+c\) A1A1
OR
\(\int \frac{\textup{tan}(\textup{In}\, y)}{y}\textup{d}y=\textup{In}\left | \textup{sec}(\textup{In}\, y) \right |+c\) A1A1 [6]
Extra question
\(\textup{In}\left | \textup{sin}(\textup{In}\, y) \right |+c\)
Question
Using the substitution \(2x=\textup{sin}\theta \), or otherwise, find \(\int \left ( \sqrt{1-4x^{2}} \right )dx\). (Total 6 marks)
▶️Answer/Explanation
Ans
\(\int \left (\sqrt{1-4x^{2}} \right )\textup{d}x\)
\(\textup{Let}\, 2x=\textup{sin}\theta \Rightarrow 2\frac{\textup{d}x}{\textup{d}\theta }=\textup{cos}\, \theta \Rightarrow \textup{d}x=\frac{1}{2}\textup{cos}\, \theta \, \textup{d}\, \theta\)
\(\Rightarrow \int \left ( \sqrt{1-4x^{2}} \right )\textup{d}x=\int \sqrt{1-\textup{sin}^{2}\theta \frac{1}{2}}\textup{cos}\,\theta \textup{d}\, \theta \)
\(=\int \frac{1}{2}\textup{cos}^{2}\theta \, \textup{d}\, \theta \) (A1)
\(\int \frac{1}{4}(\textup{cos}\, 2\theta +1)\textup{d}\theta \) (A1)
\(=\frac{1}{8}\textup{sin}\, 2\theta +\frac{\theta }{4}+C\) (A1)(A1)
\(=\frac{1}{4}\left [ 2x\sqrt{1-4x^{2}}+\textup{arcsin}\, 2x \right ]+C\) (A1)(A1) (C6) [6]