Mathematics

# If ${ I }_{ n }=\int { \sin ^{ n }{ x } dx },$, then ${ nI }_{ n }-\left( n-1 \right) { I }_{ n-2 }=$

$-\sin ^{ n-1 }{ x\cos { x } }$

##### SOLUTION
${I_n} = \int_{}^{} {{{\sin }^n}xdx}$
$= \int_{}^{} {{{\sin }^{n - 1}}x\sin xdx}$
$= - {\sin ^{n - 1}}x\cos x - \int_{}^{} {\left( {n - 1} \right){{\sin }^{n - 2}}x\cos x \times - \cos xdx}$
$= - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right)\int_{}^{} {{{\sin }^{n - 2}}x\left( {1 - {{\sin }^2}x} \right)dx}$
$\ = - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right)\int_{}^{} {\left( {{{\sin }^{n - 2}}x - {{\sin }^n}x} \right)dx}$
$= - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right)\int_{}^{} {{{\sin }^{n - 2}}xdx - \left( {n - 1} \right)\int_{}^{} {{{\sin }^n}xdx} }$
$\Rightarrow {I_n} = - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right){I_{n - 2}} - \left( {n - 1} \right){I_n}$
$\Rightarrow n{I_n} = - {\sin ^{n - 1}}x\cos x + \left( {n - 1} \right){I_{n - 2}}$
$\Rightarrow n{I_n} - \left( {n - 1} \right){I_{n - 2}} = - {\sin ^{n - 1}}x\cos x$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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