int_{0}^{infty} e^{-x} sin^{6} x dx = | vedclass

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(D) The general formula for the integral $I_{m} = \int_{0}^{\infty} e^{-x} \sin^{m} x dx$ is given by $I_{m} = \frac{m(m - 1)}{1 + m^{2}} I_{m - 2}$ f

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$\frac{144}{629}$

Vedclass · Answer

(D) The general formula for the integral $I_{m} = \int_{0}^{\infty} e^{-x} \sin^{m} x dx$ is given by $I_{m} = \frac{m(m - 1)}{1 + m^{2}} I_{m - 2}$ for $m > 2$. Applying this recurrence relation for $m = 6$: $I_{6} = \frac{6(5)}{1 + 6^{2}} I_{4} = \frac{30}{37} I_{4}$ $I_{4} = \frac{4(3)}{1 + 4^{2}} I_{2} = \frac{12}{17} I_{2}$ $I_{2} = \frac{2(1)}{1 + 2^{2}} I_{0} = \frac{2}{5} I_{0}$ Now,calculate $I_{0}$: $I_{0} = \int_{0}^{\infty} e^{-x} dx = [-e^{-x}]_{0}^{\infty} = 0 - (-1) = 1$ Substituting these values back: $I_{6} = \frac{30}{37} 	imes \frac{12}{17} 	imes \frac{2}{5} 	imes 1 = \frac{720}{3145} = \frac{144}{629}$.

$\int_{0}^{\infty} e^{-x} \sin^{6} x dx =$

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