Let ${I_n} = \mathop \smallint \limits_0^{\pi /2} \,\,{\sin ^n}\,\,x\,\,dx,\,\,n \in N$. Then

${I_n}:{I_{n - 2}} = n:\left( {n - 1} \right)$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Analyzing +4 | -1

Let

{I_n} = \int \limits_0^{\pi /2} \,\,{\sin ^n}\,\,x\,\,dx,\,\,n \in N

. Then

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it represents the correct reduction formula derived from integration by parts.

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