2003 Paper 3 Q1

Year: 2003
Paper: 3
Question Number: 1

Course: UFM Pure
Section: Integration using inverse trig and hyperbolic functions

Difficulty: 1700.0 Banger: 1516.0

integration inverse trigonometric functions inverse hyperbolic functions arcsin arcosh chain rule differentiation algebraic manipulation

Problem

Given that $x+a>0$ and $x+b>0\,$, and that $b>a\,$, show that \[ \frac{\mathrm{d} \ }{\mathrm{d} x} \arcsin \left ( \frac{x + a }{ \ x + b} \right) = \frac{ \sqrt{\;b - a\;}} {( x + b ) \sqrt{ a + b + 2x} \ \ } \] and find $\displaystyle \frac{\mathrm{d} \ }{ \mathrm{d} x} \; \mathrm{arcosh} \left ( \frac{x + b }{ \ x + a} \right)$. Hence, or otherwise, integrate, for $x > -1\,$,

\[ \int \frac{1}{ ( x + 1) \sqrt{x + 3} } \mathrm{d} x \]
\[ \int \frac{1} {( x + 3 ) \sqrt{x + 1} } \mathrm{d} x \] .

[You may use the results $\frac{\d \ }{\d x} \arcsin x = \frac 1 {\sqrt{1-x^2\;}\;}$ and $ \frac{\d \ }{\d x} \; {\rm arcosh } \; x = \frac 1 {\sqrt{x^2-1}\;}\;$. ]

Solution

\begin{align*} \frac{\mathrm{d} \ }{\mathrm{d} x} \arcsin \left ( \frac{x + a }{ \ x + b} \right ) &= \frac{1}{\sqrt{1-\left ( \frac{x + a }{ \ x + b} \right )^2}} \left ( \frac{b-a}{(x+b)^2} \right) \\ &= \frac{b-a}{(x+b)\sqrt{(x+b)^2-(x+a)^2}} \\ &= \frac{b-a}{(x+b)\sqrt{(b-a)(2x+b+a)}} \\ &= \frac{\sqrt{b-a}}{(x+b)\sqrt{a+b+2x}} \\ \\ \frac{\mathrm{d} \ }{ \mathrm{d} x} \; \mathrm{arcosh} \left ( \frac{x + b }{ \ x + a} \right) &= \frac{1}{\sqrt{\left ( \frac{x + b }{ \ x + a} \right)^2-1}} \left ( -\frac{b-a}{(x+a)^2} \right) \\ &= -\frac{b-a}{(x+a)\sqrt{(x+b)^2-(x+a)^2}} \\ &= -\frac{b-a}{(x+a)\sqrt{(b-a)(a+b+2x)}} \\ &= -\frac{\sqrt{b-a}}{(x+a)\sqrt{a+b+2x}} \end{align*}

\begin{align*} \int \frac{1}{ ( x + 1) \sqrt{x + 3} } \mathrm{d} x &= \int \frac{1}{(x+1)\sqrt{\frac12 (2x+6)}} \d x\\ &= \int \frac{\sqrt{2}}{(x+1)\sqrt{2x+1+5}} \d x \\ &= \frac{\sqrt{2}}{2}\int \frac{\sqrt{5-1}}{(x+1)\sqrt{2x+1+5}} \d x \\ &= - \frac{\sqrt{2}}{2}\textrm{arcosh} \left ( \frac{x+5}{x+1} \right) + C \end{align*}
\begin{align*} \int \frac{1}{(x+3)\sqrt{x+1}} \d x &= \int \frac{1}{(x+3)\sqrt{\tfrac12(2x+2)}} \d x + C \\ &= \int \frac{\sqrt{3-1}}{(x+3)\sqrt{2x+3-1}} \d x \\ &= \textrm{arcsin} \left ( \frac{x-1}{x+3} \right) \end{align*}

Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1516.0

Banger Comparisons: 1

Show LaTeX source

Problem source

Given that $x+a>0$ and $x+b>0\,$, and that $b>a\,$, show that   
\[  
\frac{\mathrm{d} \ }{\mathrm{d} x}   
\arcsin \left ( \frac{x + a }{ \ x + b} \right) =   
\frac{ \sqrt{\;b - a\;}} {( x + b ) \sqrt{ a + b + 2x} \ \ }  
\]  
and find $\displaystyle   
\frac{\mathrm{d} \ }{  \mathrm{d} x} \;  \mathrm{arcosh} \left ( \frac{x + b }{ \ x + a} \right)$.  
  
Hence, or otherwise, integrate, for $x > -1\,$,  
\begin{questionparts}  
\item \[ \int \frac{1}{ ( x + 1) \sqrt{x + 3} }  \mathrm{d} x \] 
\item \[ \int \frac{1} {( x + 3 ) \sqrt{x + 1} } \mathrm{d} x \] .  
\end{questionparts}  
  
[You may use the results 
$\frac{\d \ }{\d x} \arcsin x = \frac 1 {\sqrt{1-x^2\;}\;}$  and  $ \frac{\d \ }{\d x} \; {\rm arcosh } \; x = \frac 1 {\sqrt{x^2-1}\;}\;$. ]

Solution source

\begin{align*}
\frac{\mathrm{d} \ }{\mathrm{d} x}   
\arcsin \left ( \frac{x + a }{ \ x + b} \right ) &= \frac{1}{\sqrt{1-\left ( \frac{x + a }{ \ x + b} \right )^2}} \left ( \frac{b-a}{(x+b)^2} \right) \\
&= \frac{b-a}{(x+b)\sqrt{(x+b)^2-(x+a)^2}} \\
&= \frac{b-a}{(x+b)\sqrt{(b-a)(2x+b+a)}} \\
&= \frac{\sqrt{b-a}}{(x+b)\sqrt{a+b+2x}} \\
\\
\frac{\mathrm{d} \ }{  \mathrm{d} x} \;  \mathrm{arcosh} \left ( \frac{x + b }{ \ x + a} \right) &= \frac{1}{\sqrt{\left ( \frac{x + b }{ \ x + a} \right)^2-1}} \left ( -\frac{b-a}{(x+a)^2} \right) \\
&= -\frac{b-a}{(x+a)\sqrt{(x+b)^2-(x+a)^2}} \\
&= -\frac{b-a}{(x+a)\sqrt{(b-a)(a+b+2x)}} \\
&= -\frac{\sqrt{b-a}}{(x+a)\sqrt{a+b+2x}}
\end{align*}

\begin{questionparts}
\item \begin{align*}
\int
\frac{1}{ ( x + 1) \sqrt{x + 3} }  \mathrm{d} x &= \int \frac{1}{(x+1)\sqrt{\frac12 (2x+6)}} \d x\\
&= \int \frac{\sqrt{2}}{(x+1)\sqrt{2x+1+5}} \d x \\
&= \frac{\sqrt{2}}{2}\int \frac{\sqrt{5-1}}{(x+1)\sqrt{2x+1+5}} \d x \\
&= - \frac{\sqrt{2}}{2}\textrm{arcosh} \left ( \frac{x+5}{x+1} \right) + C
\end{align*}
\item \begin{align*}
\int \frac{1}{(x+3)\sqrt{x+1}} \d x &= \int \frac{1}{(x+3)\sqrt{\tfrac12(2x+2)}} \d x + C \\
&= \int \frac{\sqrt{3-1}}{(x+3)\sqrt{2x+3-1}} \d x \\
&=  \textrm{arcsin} \left ( \frac{x-1}{x+3} \right)
\end{align*}
\end{questionparts}

Back to List