Mathematics

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


ANSWER

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


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
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