Let I n -0-2sin 2 n -1 x -sin x dx - J n -0-2sin 2 n x -sin 2 x dx - n - N- thenA- Jn-1-Jn-InB- Jn-1-Jn-In-1C- J n -1- J n - J nD- J n -1- J n -1- J n

The correct option is B J_(n+1) J_n=I_(n+1)J_n J_n 1=∫_0^π/2sin^2 nx sin^2(n 1)x/sin^2 xdx ∫_0^π/2sin^2(2n 1)x sin x/sin^2 xdx =I_n i.e J_n J_n 1=I_n ⇒ J_n+1 ...

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Byju's Answer

Standard XII

Mathematics

Basic Inverse Trigonometric Functions

Let I n =∫0π/...

Question

Let $I_{n} = \int_{0}^{\frac{π}{2}} \frac{s i n (2 n - 1) x}{s i n x} d x, J_{n} = \int_{0}^{\frac{π}{2}} \frac{s i n^{2} n x}{s i n^{2} x} d x, n \in N$ , then

J_{(n + 1)} - J_{n} = I_{n}

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J_{(n + 1)} - J_{n} = I_{(n + 1)}

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J_{n + 1} + J_{n} = J_{n}

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J_{n + 1} + J_{n + 1} = J_{n}

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Solution

The correct option is B $J_{(n + 1)} - J_{n} = I_{(n + 1)}$
$J_{n} - J_{n - 1} = \int_{0}^{\frac{π}{2}} \frac{s i n^{2} n x - s i n^{2} (n - 1) x}{s i n^{2} x} d x - \int_{0}^{\frac{π}{2}} \frac{s i n^{2} (2 n - 1) x s i n x}{s i n^{2} x} d x = I_{n}$
i.e $J_{n} - J_{n - 1} = I_{n} \Rightarrow J_{n + 1} - J_{n} = I_{n + 1}$

Suggest Corrections

Let In=∫π20sin(2n−1)xsinxdx,Jn=∫π20sin2nxsin2xdx,n∈N, then

The correct option is B J(n+1)−Jn=I(n+1)Jn−Jn−1=∫π20sin2nx−sin2(n−1)xsin2xdx−∫π20sin2(2n−1)x sinxsin2xdx=In i.e Jn−Jn−1=In⇒Jn+1−Jn=In+1

Let $I_{n} = \int_{0}^{\frac{π}{2}} \frac{s i n (2 n - 1) x}{s i n x} d x, J_{n} = \int_{0}^{\frac{π}{2}} \frac{s i n^{2} n x}{s i n^{2} x} d x, n \in N$ , then