2
$\begingroup$

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?

$\endgroup$

marked as duplicate by Hans Lundmark, Hakim, J. W. Perry, Yiorgos S. Smyrlis, user147263 Oct 17 '14 at 19:28

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 2
    $\begingroup$ math.stackexchange.com/questions/236543/… $\endgroup$ – lab bhattacharjee May 12 '14 at 19:05
  • 1
    $\begingroup$ Split 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. $\endgroup$ – Papagon May 12 '14 at 19:08
  • $\begingroup$ I'll just try that now - Thanks $\endgroup$ – Jbarrell May 12 '14 at 19:08
0
$\begingroup$

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 $$ \left((\cos x)^n\right)' = -n(\cos x)^{n-1}\sin x \text{.} $$

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – Jbarrell May 12 '14 at 19:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.