Question

want to learn reduction formula

Answer 1

yes

Answer 2

You have to be more specific.
You posted this in Diff EQ section so I'm going to guess you mean the Reduction of Order formula?

Answer 3

\[\int\limits_{0}^{pie/2?}\sin^nx dx\]

Answer 4

Oh oh i see :) Reduction Formula for Sine

Answer 5

\[\int sin^nx dx\]\[=\int sin^{n-1}x(sinx)dx\]\[=-\int sin^{n-1}xd(cosx)\]\[=-[sin^{n-1}xcosx-\int cosxd(sin^{n-1}x)]\]\[=-[sin^{n-1}xcosx-(n-1)\int cos^2xsin^{n-2}xdx]\]\[=-[sin^{n-1}xcosx-(n-1)\int (1-sin^2x)sin^{n-2}xdx]\]\[=-[sin^{n-1}xcosx-(n-1)\int (sin^{n-2}x-sin^{n}x)dx]\]\[=-sin^{n-1}xcosx+(n-1)\int (sin^{n-2}x-sin^{n}x)dx\]\[=-sin^{n-1}xcosx+(n-1)\int sin^{n-2}xdx-(n-1)\int sin^{n}xdx\]So,
\[\int sin^nx dx=-sin^{n-1}xcosx+(n-1)\int sin^{n-2}xdx-(n-1)\int sin^{n}xdx\]\[n\int sin^nx dx = -sin^{n-1}xcosx+(n-1)\int sin^{n-2}xdx\]\[\int sin^nx dx = -\frac{1}{n}sin^{n-1}xcosx+\frac{(n-1)}{n}\int sin^{n-2}xdx\]