Im,n∫sinmxcosnxdx=sinm+1xcosn−1m+n+n−1m+nIm,n−2m=5,n=4=−sinm−1⁡xcosn+1⁡xm+n+m−1m+nIm−2,n(m,n∈N,m≥2,n≥2)I5=∫sin5⁡xcos4⁡xdx=−sin4⁡xcos5⁡x9+49I3,4=−sin4⁡xcos5⁡x9+49−sin2⁡xcos5⁡x7+27I1,4=−sin4⁡xcos5⁡x9−463sin2⁡xcos5⁡x+863∫sin⁡xcos4⁡xdx=−sin4⁡xcos5⁡x9−463sin2⁡xcos5⁡x−8cos5⁡x315+C

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$\int \sin^{5} x \cos^{4} x d x =$

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$\frac{\sin^{4} x \cos^{5} x}{9} + \frac{4 \sin^{2} x \cos^{5} x}{63} + \frac{\cos^{5} x}{315} + C$

$\frac{\sin^{4} x \cos^{5} x}{9} - \frac{4 \sin^{2} x \cos^{5} x}{63} + \frac{\cos^{5} x}{315} + C$

$- \frac{\sin^{4} x \cos^{5} x}{9} + \frac{4 \sin^{2} x \cos^{5} x}{63} + \frac{\cos^{5} x}{315} + C$

$- \frac{\sin^{4} x \cos^{5} x}{9} - \frac{4 \sin^{2} x \cos^{5} x}{63} - \frac{8 \cos^{5} x}{315} + C$

answer is D.

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Detailed Solution

$I_{m, n} \int \sin^{m} x \cos^{n} x d x = \frac{\sin^{m + 1} x \cos^{n - 1}}{m + n} + \frac{n - 1}{m + n} I_{m, n - 2}$

$m = 5, n = 4$

$\begin{array}{l} = \frac{- \sin^{m - 1} x \cos^{n + 1} x}{m + n} + \frac{m - 1}{m + n} I_{m - 2, n} (m, n \in N, m \geq 2, n \geq 2) \\ I_{5} = \int \sin^{5} x \cos^{4} x d x = \frac{- \sin^{4} x \cos^{5} x}{9} + \frac{4}{9} I_{3, 4} \\ = \frac{- \sin^{4} x \cos^{5} x}{9} + \frac{4}{9} [\frac{- \sin^{2} x \cos^{5} x}{7} + \frac{2}{7} I_{1, 4}] \\ = \frac{- \sin^{4} x \cos^{5} x}{9} - \frac{4}{63} \sin^{2} x \cos^{5} x + \frac{8}{63} \int \sin x \cos^{4} x d x \\ = \frac{- \sin^{4} x \cos^{5} x}{9} - \frac{4}{63} \sin^{2} x \cos^{5} x - \frac{8 \cos^{5} x}{315} + C \end{array}$