int_0^{pi / 2} sin ^m x cos ^4 x , dx = frac{ | vedclass

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(A) We use the Wallis formula for the definite integral: $\int_0^{\pi / 2} \sin ^m x \cos ^n x \, dx = \frac{[(m-1)(m-3)\dots] \cdot [(n-1)(n-3)\dots]

Vedclass · Accepted Answer

$8$

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(A) We use the Wallis formula for the definite integral: $\int_0^{\pi / 2} \sin ^m x \cos ^n x \, dx = \frac{[(m-1)(m-3)\dots] \cdot [(n-1)(n-3)\dots]}{(m+n)(m+n-2)\dots} \cdot k$,where $k = \frac{\pi}{2}$ if both $m$ and $n$ are even,and $k = 1$ otherwise.Given $\int_0^{\pi / 2} \sin ^m x \cos ^4 x \, dx = \frac{7 \pi}{2048}$.Since the result contains $\pi$,both $m$ and $4$ must be even,so $m$ is even and $k = \frac{\pi}{2}$.Substituting $n = 4$ into the formula:$I = \frac{(m-1)(m-3)\dots(1) \cdot (4-1)(4-3)}{(m+4)(m+2)(m)(m-2)\dots(2)} \cdot \frac{\pi}{2} = \frac{7 \pi}{2048}$.For $m = 8$:$I = \frac{(7 \cdot 5 \cdot 3 \cdot 1) \cdot (3 \cdot 1)}{(12 \cdot 10 \cdot 8 \cdot 6 \cdot 4 \cdot 2)} \cdot \frac{\pi}{2} = \frac{105 \cdot 3}{46080} \cdot \frac{\pi}{2} = \frac{315 \pi}{92160} = \frac{7 \pi}{2048}$.Thus,$m = 8$.

$\int_0^{\pi / 2} \sin ^m x \cos ^4 x \, dx = \frac{7 \pi}{2048} \Rightarrow m = ?$

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