Question

$\tt{ I ​​_{​n} = \: \int\limits_{​}^{​}​ {tan}^{x} ​\,​ dx } \\ \tt{I ​​_{​n} = \frac{ {(tan \: x)}^{n - 1} }{n - 1} - I ​​_{​n - 2} } \: \: (prove \: it)$
​

Answer 1

$\large{\underline{\underline{ \bf{⎆Correct\:Question:-}}}}$

❐ Given :

$\tt{ I _{n} = \: \int\limits_{}^{} {tan}^{n} x\, dx }$

❐ To prove :

$\tt{I _{n} = \frac{ {(tan \: x)}^{n - 1} }{n - 1} - I _{n - 2} } \: \:$

$\large{\underline{\underline{ \bf{☂ Solution:-}}}}$

$\longmapsto \: \tt{ I _{n} = \: \int\limits {tan}^{x} \, dx }$

$\longmapsto \tt{ I _{n} = {\int\limits(tan}^{n - 2} . {tan}^{2} x)dx}$

$\longmapsto \tt{ I _{n} = {\int\limits(tan}^{n - 2}x . ({sec}^{2}x - 1 ))dx}$

$\longmapsto\: \tt{ I _{n} = {\int\limits(tan}^{n - 2} x \: {sec}^{2}x - {tan}^{n - 2}x )dx}$

$\longmapsto \tt{ I _{n} = {\int\limits(tan}^{n - 2} x \: {sec}^{2}x )\: dx - \int\limits{(tan}^{n - 2}x )dx}$

$➥ \: \tt{Put , \:tan\:x\:=\:t}$

$\tt{After\:differentiating\: both\:the\: sides\:,\:we \: get - }$

$➥ \: \tt{sec²x\:dx=dt}$

$\therefore\tt{ I _{n} \: = \int\limits( {t}^{n - 2})dt - I _{n - 2}}$

$\tt{Integrate\:both\:the\:sides - }$

$\longmapsto \: \tt{ I _{n} \: = ( \frac{{t}^{n - 1}}{n - 1} ) - I _{n - 2}}$

$\tt{Put\:the\: value\:of \: 't' \::}$

$\:✰ \: \tt{ I _{n} \: = ( \frac{{tan}^{n - 1}}{n - 1} ) - I _{n - 2}}$

• Hence proved .