If In=∫cosnxdx, then 6I6−5I4=

=∫cosn−1×cosxdx I II cosn−1x.sinx−∫(n−1)cosn−2x(−sinx)(sinx)dx cosn−1xsinx+∫cosn−2x.(n−1)sin2xdx =cosn−1xsinx+(n−1)∫cosn−2x−(n−1)∫cosnxdx=cosn−1xsinx+(n−1)In−2−(n−1)In⇒nIn−(n−I)In−2=cosn−1nsinxPutting n = 6 we get6I6−5I4=cos5xsinx

If In=∫cosnxdx, then 6I6−5I4=

=∫cosn−1×cosxdx I II cosn−1x.sinx−∫(n−1)cosn−2x(−sinx)(sinx)dx cosn−1xsinx+∫cosn−2x.(n−1)sin2xdx =cosn−1xsinx+(n−1)∫cosn−2x−(n−1)∫cosnxdx=cosn−1xsinx+(n−1)In−2−(n−1)In⇒nIn−(n−I)In−2=cosn−1nsinxPutting n = 6 we get6I6−5I4=cos5xsinx

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If $I_{n} = \int \cos^{n} x d x,$ then $6 I_{6} - 5 I_{4} =$

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$- \cos^{5} x \sin^{2} x$

$\cos^{6} x \sin^{2} x$

$\cos^{3} x \sin^{2} x$

$\cos^{5} x \sin x$

answer is D.

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Detailed Solution

$= \underset{I I I}{\int \cos^{n - 1} \times \cos x d x}$
$\cos^{n - 1} x . \sin x - \int (n - 1) \cos^{n - 2} x (- \sin x) (\sin x) d x \cos^{n - 1} x \sin x + \int \cos^{n - 2} x . (n - 1) \sin^{2} x d x \begin{array}{l} = \cos^{n - 1} x \sin x + (n - 1) \int \cos^{n - 2} x - (n - 1) \int \cos^{n} x d x \\ = \cos^{n - 1} x \sin x + (n - 1) I_{n - 2} - (n - 1) I_{n} \\ \Rightarrow n I_{n} - (n - I) I_{n - 2} = \cos^{n - 1} n \sin x \end{array}$
Putting n = 6 we get
$6 I_{6} - 5 I_{4} = \cos^{5} x \sin x$