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Q.
If $\int \cos ^{ n } x \sin x dx =-\frac{\cos ^{6} x }{6}+ C$, then $n =$
Integrals
Solution:
Let $I =\int \cos ^{ n } x \sin x d x$
Put $\cos x = t$
$-\sin x dx = dt$
$\therefore I =-\int t ^{ n } dt =-\frac{ t ^{n +1}}{ n +1}+ C$
$=-\frac{\cos ^{ n +1} x }{ n +1}+ C$
$=-\frac{\cos ^{6} x }{6}+ C$
$\therefore n +1=6$ or $n =5$