Question

The Integral from 2 to sqrt2 of 1/(t^3sqrt(t^2-1))dt

Answer 1

\[\int\limits_{2}^{\sqrt{2}}\frac{ 1 }{ t^3\sqrt{t^2-1} }dt\] ?

Answer 2

Just the integral of 2 is on top.

Answer 3

what ?

Answer 4

reverse 2 and sqrt(2)

Answer 5

ok

Answer 6

?

Answer 7

ok well it took a while but :

\[t=coshx\]
\[dt=sinhxdx\]

then your integral become :
\[\int\limits_{\sqrt2}^{2} \frac{ 1 }{ \cosh^3x }dx\]

now using integration by parts
\[\int\limits_{\sqrt2}^{2} \frac{ 1 }{ coshx }\frac{ 1 }{ \cosh^2x }dx\]

\[v=\frac{ 1 }{ coshx } , u'=\frac{ 1 }{ \cosh^2x } \]

Answer 8

\[v'=\frac{ -sinhx }{ \cosh^2x } , u=tanhx\]
so we get by integration by parts :

\[\int\limits_{}^{}\frac{ 1 }{ \cosh^3x }=\frac{ tanhx }{ coshx } + \int\limits_{}^{}\frac{ \sinh^2x }{ \cosh^3x }\]
we can write it as :

\[\int\limits_{}^{}\frac{ 1 }{ \cosh^3x }=\frac{ tanhx }{ coshx } + \int\limits_{}^{}\frac{ \cosh^2x-1 }{ \cosh^3x }\]

then we have:
\[2\int\limits_{}^{}\frac{ 1 }{ \cosh^3x }=\frac{ tanhx }{ coshx } + \int\limits_{}^{}\frac{ 1 }{ coshx }\]

we divide by 2 and in order to solve 1/coshx i do :
\[\int\limits_{}^{}\frac{ 1 }{ \cosh^3x }=\frac{ tanhx }{ 2coshx } + \frac{ 1 }{ 2 } \int\limits_{}^{}\frac{ 1 }{ coshx }\frac{ coshx }{ coshx }\]

the n we get:

\[\int\limits_{}^{}\frac{ 1 }{ \cosh^3x }=\frac{ tanhx }{ 2coshx } + \frac{ 1 }{ 2 } \int\limits_{}^{}\frac{ coshx }{ 1+\sinh^2x }\]

in order to solve

Answer 9

\[\int\limits_{}^{}\frac{ coshx }{ 1+\sinh^2x }\]
i use another sub p = sinhx so dp=coshxdx

then this integral is easy and i get arctan(p)

so we have \[\int\limits_{\sqrt2}^{2}\frac{ 1 }{ \cosh^3x } = \frac{ tanhx }{ 2coshx } + \frac{ 1 }{ 2 }\arctan(sinhx)\]

now you can answer remembering coshx = t