If ${I_n} = \smallint {\sin ^n}\,\,x\,dx$,then $n{I_n}\,\, - \,\,\left( {n - 1} \right){I_{n - 2}}$ equals

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

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

JEE MATHEMATICS MCQ: INDEFINITE INTEGRATION

1 Easy

1 of 1

Application +4 | -1

{I_n} = \int {\sin ^n}\,\,x\,dx

,then

n{I_n}\,\, - \,\,\left( {n - 1} \right){I_{n - 2}}

equals

View answer and explanation

Correct option: C

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

Option C is correct because it is the result of applying integration by parts to the integral of sin^n(x).

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