2004 Paper 3 Q1

Year: 2004
Paper: 3
Question Number: 1

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

Difficulty: 1700.0 Banger: 1603.9
integration hyperbolic functions inverse trigonometric functions partial fractions substitution improper integral logarithmic result definite integral

Problem

Show that \[ \int_0^a \frac{\sinh x}{2\cosh^2 x -1} \, \mathrm{d} x = \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}\cosh a -1}{\sqrt{2}\cosh a +1}\r + \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \] and find \[ \int_0^a \frac{\cosh x}{1+2\sinh^2 x} \, \mathrm{d} x \, . \] Hence show that \[ \int_0^\infty \frac{\cosh x - \sinh x}{1+2\sinh^2 x} \, \mathrm{d} x = \frac{\pi}{2\sqrt{2}} - \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \, . \] By substituting \(u = \e^x\) in this result, or otherwise, find \[ \int_1^\infty \frac{1}{1+u^4} \, \mathrm{d} u \, . \]

No solution available for this problem.

Rating Information

Difficulty Rating: 1700.0

Difficulty Comparisons: 0

Banger Rating: 1603.9

Banger Comparisons: 10

Show LaTeX source
Problem source
Show that
\[
\int_0^a \frac{\sinh x}{2\cosh^2 x -1} \, \mathrm{d} x = \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}\cosh a -1}{\sqrt{2}\cosh a +1}\r + \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r
\]
and find
\[
\int_0^a \frac{\cosh x}{1+2\sinh^2 x} \, \mathrm{d} x \, .
\]
Hence show that 
\[
\int_0^\infty \frac{\cosh x - \sinh x}{1+2\sinh^2 x} \, \mathrm{d} x = \frac{\pi}{2\sqrt{2}} - \frac{1}{2 \sqrt{2}} \ln \l \frac{\sqrt{2}+1}{\sqrt{2}-1}\r \, .
\]
By substituting $u = \e^x$ in this result, or otherwise, find
\[
\int_1^\infty \frac{1}{1+u^4} \, \mathrm{d} u \, .
\]
Back to List