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If \(I_n=\int_0^{\pi / 2} \sin ^n(x) d x\) and \(I_n=(k) I_{n-2}\), then what will be the value of \(k\) ?
Options:
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Verified Answer
The correct answer is:
\(\frac{n-1}{n}\)
\(\begin{aligned}
\mathrm{I}_n & =\int_0^{\pi / 2} \sin x \cdot \sin ^{n-1} x \mathrm{dx} \\
& \left.=\sin ^{n-1} x(-\cos x)\right]_0^{\pi / 2}+\int_0^{\pi / 2}(n-1) \cdot \sin ^{n-2} x \\
& =0+(n-1) \int_0^{\pi / 2} \sin ^{n-2} x\left(1-\sin ^2 x\right) d x \\
\mathrm{I}_n & =(n-1) \int_0^{\pi / 2} \sin ^{n-2} x d x-(n-1) \int_0^{\pi / 2} \sin _x^n d x \\
n \mathrm{I}_n & =(n-1) \cdot \mathrm{I}_{n-2} \\
\mathrm{I}_n & =\frac{n-1}{n} \cdot \mathrm{I}_{n-2}
\end{aligned}\)
Hence, option (b) is correct.
\mathrm{I}_n & =\int_0^{\pi / 2} \sin x \cdot \sin ^{n-1} x \mathrm{dx} \\
& \left.=\sin ^{n-1} x(-\cos x)\right]_0^{\pi / 2}+\int_0^{\pi / 2}(n-1) \cdot \sin ^{n-2} x \\
& =0+(n-1) \int_0^{\pi / 2} \sin ^{n-2} x\left(1-\sin ^2 x\right) d x \\
\mathrm{I}_n & =(n-1) \int_0^{\pi / 2} \sin ^{n-2} x d x-(n-1) \int_0^{\pi / 2} \sin _x^n d x \\
n \mathrm{I}_n & =(n-1) \cdot \mathrm{I}_{n-2} \\
\mathrm{I}_n & =\frac{n-1}{n} \cdot \mathrm{I}_{n-2}
\end{aligned}\)
Hence, option (b) is correct.
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