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If $I_n=\int \sin ^n x dx $, then $n\; I_n-(n-1)I_{n-2}$=

\[\begin {array} {1 1} (1)\;\sin ^{n-1} x \cos x & \quad (2)\;\cos ^{n-1} x \sin x \\ (3)\;-\sin ^{n-1}x \cos x & \quad (4)\;-\cos ^{n-1}x \sin x \end {array}\]

1 Answer

$ (3)\;-\sin ^{n-1}x \cos x$
answered Nov 7, 2013 by pady_1
 
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