Question

$\displaystyle{\rm \int _0^{log\left({1+\sqrt{2}}\right)}\:\left(\frac{e^x-e^{-x}}{2}\right)^3\cdot \left(\frac{e^x + e^{-x}}{2}\right)^{11}\:dx}$​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\displaystyle{\rm \int _0^{log\left({1+\sqrt{2}}\right)}\:\left(\frac{e^x-e^{-x}}{2}\right)^3\cdot \left(\frac{e^x + e^{-x}}{2}\right)^{11}\:dx}$

can be rewritten as

$\rm \:=\dfrac{1}{ {2}^{14} } \displaystyle{\rm \int _0^{log\left({1+\sqrt{2}}\right)}\: {( {e}^{x} + {e}^{ - x} )}^{11} {({e}^{x} - {e}^{ - x})}^{3} \:dx}$

can be further rewritten as

$\rm \:=\dfrac{1}{ {2}^{14} } \displaystyle{\rm \int _0^{log\left({1+\sqrt{2}}\right)}\: {( {e}^{x} + {e}^{ - x} )}^{11} {({e}^{x} - {e}^{ - x})}^{2}({e}^{x} - {e}^{ - x}) \:dx}$

can be further rewritten as

$\rm \:=\dfrac{1}{ {2}^{14} } \displaystyle{\rm \int _0^{log\left({1+\sqrt{2}}\right)}\: {( {e}^{x} + {e}^{ - x} )}^{11} [{({e}^{x} + {e}^{ - x})}^{2} - 4]({e}^{x} - {e}^{ - x}) \:dx}$

Now, to evaluate this integral, we use method of Substitution.

So, substitute

$\rm \: {e}^{x} + {e}^{ - x} = y$

$\rm \: ({e}^{x} - {e}^{ - x})dx = dy$

We know, when we substitute in definite integral, we have to change the limits too.

$\red{\rm \: When \: x = 0}$

$\rm \: {e}^{0} + {e}^{ - 0} = y\rm\implies \:y = 2$

$\red{\rm \: When \: x = log(1 + \sqrt{2} ) }$

$\rm \: y = {e}^{ log(\sqrt{2} + 1) } + {e}^{ - log( \sqrt{2} + 1) }$

$\rm \: y = \sqrt{2} + 1 + \frac{1}{ \sqrt{2} + 1}$

$\rm \: y = \sqrt{2} + 1 + \sqrt{2} - 1$

$\rm \: y = 2\sqrt{2}$

So, above integral can be rewritten as

$\rm \: = \: \dfrac{1}{ {2}^{14} } \displaystyle\int_{2}^{2 \sqrt{2} } {y}^{11}[ {y}^{2} - 4] \: dy$

$\rm \: = \: \dfrac{1}{ {2}^{14} } \displaystyle\int_{2}^{2 \sqrt{2} } [{y}^{13} - 4{y}^{11} ] \: dy$

$\rm \: = \: \dfrac{1}{ {2}^{14} } \bigg(\dfrac{ {y}^{14} }{14} - \dfrac{4 {y}^{12} }{12} \bigg) _{2}^{2 \sqrt{2} }$

$\rm \: = \: \dfrac{1}{ {2}^{14} } \bigg(\dfrac{ {y}^{14} }{14} - \dfrac{ {y}^{12} }{3} \bigg) _{2}^{2 \sqrt{2} }$

$\rm \: = \: \dfrac{1}{ {2}^{14} } \bigg(\dfrac{ {2}^{21} - {2}^{14} }{14} - \dfrac{ {2}^{18} - {2}^{12} }{3} \bigg)$

$\rm \: = \: \dfrac{1}{ {2}^{14} } \bigg(\dfrac{ {2}^{14}({2}^{7} - 1)}{14} - \dfrac{ {2}^{12}({2}^{6} - 1)}{3} \bigg)$

$\rm \: = \: \dfrac{({2}^{7} - 1)}{14} - \dfrac{({2}^{6} - 1)}{3 \times 4}$

$\rm \: = \: \dfrac{128 - 1}{14} - \dfrac{64 - 1}{12}$

$\rm \: = \: \dfrac{127}{14} - \dfrac{63}{12}$

$\rm \: = \: \dfrac{127}{14} - \dfrac{21}{4}$

$\rm \: = \: \dfrac{254 - 147}{28}$

$\rm \: = \: \dfrac{107}{28}$