I need to find the reduction formula for the integral of $\cos^n(x)$.
Ive split it into $\cos(x)\cos^{(n-1)}x$ in the hope of integrating by parts, but I'm unsure how to differentiate $\cos^{(n-1)}$, how should I proceed?
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2http://math.stackexchange.com/questions/236543/proving-a-reduction-formula-for-the-antiderivative-of-cosnx – lab bhattacharjee May 12 '14 at 19:05
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1Split into $\cos(x)^2\cos(x)^{n-2}$, then, using $\cos(x)^2=1-\sin(x)^2$, integrate by parts $\sin(x)\sin(x)\cos(x)^{n-2}$, it gives a recurrence relation. – Papagon May 12 '14 at 19:08
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I'll just try that now - Thanks – Jbarrell May 12 '14 at 19:08
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You use the chain rule. The function $f \,:\, x \to (\cos(x))^n$ is the concatenation $g \circ h$ of $g \,:\, x \to x^n$ and $h \,:\, x \to \cos x$, so by the chain rule $$ f' = (g \circ h)' = (g'\circ h)\cdot h' $$
Since $g'(x) = nx^{n-1}$ and $h'(x) = -\sin x$, it follows that $$ \left((\cos x)^n\right)' = -n(\cos x)^{n-1}\sin x \text{.} $$
fgp
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I think I've managed it, but what I've got disagrees with my textbook! I've got the answer to be In=(1/(n-1))sinxcos^(n-1)x+I(n-2) – Jbarrell May 12 '14 at 19:32