JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Analyzing +4 | -1

Let

I_n=\int_0^{\pi / 2} \frac{\sin (2 n-1) x}{\sin x} d x, \mathcal{F}_n=\int_0^{\pi / 2} \frac{\sin ^2 n x}{\sin ^2 x}, n \in N

, then

View answer and explanation

Correct option: B

\mathcal{F}_{n+1}-\mathcal{F}_n=I_{n+1}

Option B (

\mathcal{F}_{n+1}-\mathcal{F}_n=I_{n+1}

) is correct because it follows from the derivation of the difference between consecutive terms of

\mathcal{F}

and its relation to

I

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