1994 Paper 3 Q1

Year: 1994
Paper: 3
Question Number: 1

Course: UFM Additional Further Pure
Section: Reduction Formulae

Difficulty: 1700.0 Banger: 1516.0

integration reduction formula hyperbolic functions sech integration by parts definite integral

Problem

Calculate \[ \int_{0}^{x}\mathrm{sech}\, t\,\mathrm{d}t. \] Find the reduction formula involving \(I_{n}\) and \(I_{n-2}\), where \[ I_{n}=\int_{0}^{x}\mathrm{sech}^{n}t\,\mathrm{d}t \] and, hence or otherwise, find \(I_{5}\) and \(I_{6}.\)

Solution

\begin{align*} && \int_0^x \mathrm{sech}\, t \d t &= \int_0^x \frac{2}{e^t+e^{-t}} \d t \\ &&&= \int_0^x \frac{2e^t}{e^{2t}+1} \d t \\ &&&= \left [2 \arctan e^t \right ]_0^x \\ &&&= 2\tan^{-1}e^x- \frac{\pi}{2} \end{align*} \begin{align*} && I_n &= \int_0^x \mathrm{sech}^n\, t \d t \\ &&&= \int_0^x \mathrm{sech}^{n-2}\, t \mathrm{sech}^2\, t \d t \\ &&&= \left [ \mathrm{sech}^{n-2}\, t \cdot \tanh t\right ]_0^x - \int_0^x (n-2) \mathrm{sech}^{n-3} \, t \cdot (- \tanh t \mathrm{sech}\, t) \tanh t \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)\int_0^x \mathrm{sech}^{n-2} t \tanh^2 t \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)\int_0^x \mathrm{sech}^{n-2} t (1-\mathrm{sech}^2 \, t) \d t \\ &&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2}-(n-2)I_n \\ \Rightarrow && (n-1)I_n &= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2} \\ \Rightarrow && I_n &= \frac{1}{n-1} \left ( \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2} \right) \\ \end{align*} \begin{align*} I_1 &= 2\tan^{-1}e^x- \frac{\pi}{2} \\ I_3 &= \frac12 \left ( \mathrm{sech}\, x \cdot \tanh x+ 2\tan^{-1}e^x- \frac{\pi}{2}\right) \\ &= \frac12 \mathrm{sech}\, x \cdot \tanh x+ \tan^{-1}e^x- \frac{\pi}{4} \\ I_5 &= \frac14 \left (\mathrm{sech}^3\, x \cdot \tanh x + 3 \left ( \frac12 \mathrm{sech}\, x \cdot \tanh x+ \tan^{-1}e^x- \frac{\pi}{4} \right) \right) \\ &= \frac14 \mathrm{sech}^3\, x \cdot \tanh x +\frac38 \mathrm{sech}\, x \cdot \tanh x+\frac34 \tan^{-1}e^x- \frac{3\pi}{16} \\ \\ I_2 &= \tanh x \\ I_4 &= \frac13 \left ( \mathrm{sech}^2 x\tanh x +2\tanh x\right) \\ &= \frac13 \mathrm{sech}^2 x\tanh x +\frac23 \tanh x \\ I_6 &= \frac15 \left ( \mathrm{sech}^4 x \tanh x+4\left ( \frac13 \mathrm{sech}^2 x\tanh x +\frac23 \tanh x \right) \right) \\ &= \frac15 \mathrm{sech}^4 x \tanh x + \frac4{15}\mathrm{sech}^2 x\tanh x + \frac{8}{15} \tanh x \end{align*}

Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1516.0

Banger Comparisons: 1

Show LaTeX source

Problem source

Calculate 
\[
\int_{0}^{x}\mathrm{sech}\, t\,\mathrm{d}t.
\]
Find the reduction formula involving $I_{n}$ and $I_{n-2}$, where
\[
I_{n}=\int_{0}^{x}\mathrm{sech}^{n}t\,\mathrm{d}t
\]
and, hence or otherwise, find $I_{5}$ and $I_{6}.$

Solution source

\begin{align*}
&& \int_0^x \mathrm{sech}\, t \d t &= \int_0^x \frac{2}{e^t+e^{-t}} \d t \\
&&&=  \int_0^x \frac{2e^t}{e^{2t}+1} \d t \\
&&&= \left [2 \arctan e^t \right ]_0^x \\
&&&= 2\tan^{-1}e^x- \frac{\pi}{2} 
\end{align*}

\begin{align*}
&& I_n &= \int_0^x \mathrm{sech}^n\, t \d t \\
&&&= \int_0^x \mathrm{sech}^{n-2}\, t \mathrm{sech}^2\, t \d t \\
&&&= \left [ \mathrm{sech}^{n-2}\, t \cdot \tanh t\right ]_0^x - \int_0^x (n-2) \mathrm{sech}^{n-3} \, t \cdot (- \tanh t \mathrm{sech}\, t) \tanh t \d t \\
&&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)\int_0^x \mathrm{sech}^{n-2} t \tanh^2 t \d t \\
&&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)\int_0^x \mathrm{sech}^{n-2} t (1-\mathrm{sech}^2 \, t) \d t \\
&&&= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2}-(n-2)I_n \\
\Rightarrow && (n-1)I_n &= \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2} \\
\Rightarrow && I_n &= \frac{1}{n-1} \left ( \mathrm{sech}^{n-2}\, x \cdot \tanh x+(n-2)I_{n-2} \right) \\
\end{align*}

\begin{align*}
I_1 &= 2\tan^{-1}e^x- \frac{\pi}{2} \\
I_3 &= \frac12 \left ( \mathrm{sech}\, x \cdot \tanh x+ 2\tan^{-1}e^x- \frac{\pi}{2}\right) \\
&=  \frac12 \mathrm{sech}\, x \cdot \tanh x+ \tan^{-1}e^x- \frac{\pi}{4} \\
I_5 &= \frac14 \left (\mathrm{sech}^3\, x \cdot \tanh x + 3 \left ( \frac12 \mathrm{sech}\, x \cdot \tanh x+ \tan^{-1}e^x- \frac{\pi}{4} \right) \right) \\
&= \frac14 \mathrm{sech}^3\, x \cdot \tanh x +\frac38 \mathrm{sech}\, x \cdot \tanh x+\frac34 \tan^{-1}e^x- \frac{3\pi}{16} \\
\\
I_2 &= \tanh x \\
I_4 &= \frac13 \left ( \mathrm{sech}^2 x\tanh x +2\tanh x\right) \\
&= \frac13 \mathrm{sech}^2 x\tanh x +\frac23 \tanh x \\
I_6 &= \frac15 \left ( \mathrm{sech}^4 x \tanh x+4\left ( \frac13 \mathrm{sech}^2 x\tanh x +\frac23 \tanh x  \right) \right) \\
&= \frac15 \mathrm{sech}^4 x \tanh x + \frac4{15}\mathrm{sech}^2 x\tanh x + \frac{8}{15} \tanh x
\end{align*}

Back to List