Question

Integration of 1/x^3 sin(logx).dx

Answer 1

EXPLANATION.

$\sf \: \implies \: \int \: \dfrac{1}{ {x}^{3} } \sin( log(x) )dx$

CONCEPT TO BE USED

$\sf \: \implies \: { \underline{ \underline{using \: the \: concept \: of \: integration \: by \: parts}}} \\ \\ \sf \: \implies \: { \underline{ \underline{we \: can \: say \: as \: a \: product \: rule}}}$

$\sf \: \implies \: \dfrac{d}{dx} (f(x).g(x)) = f '(x).g(x) \: + \: f(x).g '(x) \\ \\ \sf \: \implies \: \int \: d(f(x).g(x)) = \int \: f'(x).g(x)dx \: + \: \int \: f(x).g'(x)dx \\ \\ \sf \: \implies \: { \underline{ \underline{integrate \: with \: respect \: to \: \: \: \: x}}} \\ \\ \sf \: \implies \: \boxed{\int \: f(x).g'(x)dx \: = f(x).g(x) \: - \: \int \: f'(x).g(x)dx}$

$\sf \: \implies \: { \underline{ \underline{how \: to \: select \: first \: function\: }}} \\ \\ \sf \: \implies \:{ \underline{ \underline{ILATE \:}}} \\ \\ \sf \: \implies \: I \: = inverse \: trigonometry \: function \\ \sf \: \implies \: L \: = logarithmic \: function \\ \sf \: \implies \: A \: = algebraic \:function \\ \sf \: \implies \: T \: = trigonometric \: function \\ \sf \: \implies \: E \: = exponential \: function$

$\sf \: \implies \: \int \: \dfrac{1}{ {x}^{3} } \sin( log(x) )dx \\ \\ \sf \: \implies \: let \: I \: = \int \: \dfrac{1}{ {x}^{3} } \sin( log(x) )dx \\ \\ \sf \: \implies \: log(x) = t \\ \\ \sf \: \implies \: x \: = {e}^{t} \\ \\ \sf \: \implies \: dx \: = {e}^{t} dt$

$\sf \: \implies \: I \: = \int \: \dfrac{1}{ {e}^{3t} } \sin(t) e {}^{t}dt \\ \\ \sf \: \implies \: I \: = \int \: {e}^{ - 2t} \sin(t) dt \\ \\ \sf \: \implies \: considered \: \sin(t) as \: first \: function \\ \sf \: \implies \: considered \: {e}^{ - 2t} as \: second \: function$

$\sf \: \implies \: I = \: \sin(t) ( \dfrac{e {}^{ - 2t} }{ - 2}) \: - \: \int \: \cos(t) \dfrac{ {e}^{ - 2t} }{2} dt \\ \\ \sf \: \implies \: I \: = \: \frac{ \sin(t) {e}^{ - 2t} }{ - 2} \: \: + \: \: \frac{1}{2} \int \: \cos(t) {e}^{ - 2t} dt \\ \\ \sf \: \implies \: I \: = \: \frac{ \sin(t) {e}^{ - 2t} }{ - 2} \: \: + \: \frac{1}{2} \Bigg[ \cos(t) \frac{e {}^{ - 2t} }{2} \: \: - \int ( - \sin(t) \frac{ {e}^{ - 2t} }{ - 2}dt \Bigg ]$

$\sf \: \implies \: I \: = \: \dfrac{ \sin(t) {e}^{ - 2t} }{ - 2} \: - \: \dfrac{1}{4} \cos(t) {e}^{ - 2t} \: - \: \int \: \dfrac{ {e}^{ - 2t} \sin(t)dt }{4} \\ \\ \sf \: \implies \: I = \: {e}^{ - 2t} \Bigg[ \frac{ - 2 \sin(t) \: - \: \cos(t) }{4} \Bigg] \: - \: \frac{ I }{4} \\ \\ \sf \: \implies \: \frac{5 I }{4} \: = \: {e}^{ - 2t} \Bigg[ \frac{ - 2 \sin(t) \: - \: \cos(t) }{4}\Bigg]$

$\sf \: \implies \: I \: = \: \dfrac{ {e}^{ - 2t} }{5} [ - 2 \sin(t) \: - \: \cos(t) ] \: + \: c \\ \\ \sf \: \implies \: I \: = \frac{ { - x}^{ - 2} }{5} [ \: 2 \sin( logx ) \: + \cos( logx ) ] \: + \: c \\ \\ \sf \: \implies \: \boxed{ \underline{ \underline{I \: = \: \frac{ - 1}{5 {x}^{2} } [ \cos( log(x) ) \: + \: 2 \sin( log(x) ) \: ] \: + \: c}}}$