Question

I'm solving for a partial differential equation, and I'm getting caught up with the integration step. Can someone help? The integral is listed below.

Answer 1

\[\int\limits_{0}^{\pi}(\sin x) ^2 (\sin nx) dx\]

Answer 2

\[\sin^{2}x = 1- \cos^{2}x\]
\[\cos 2 x = 2*\cos^{2} x -1\]

Answer 3

\[\frac{ \cos2x +1 }{ 2 }= \cos^{2}x\]

Answer 4

\[\int\limits_{0}^{\pi}(1-\cos^2x)(\sin nx)*dx\]

Answer 5

\[\int\limits_{0}^{\pi}(1-\frac{ \cos2x+1 }{ 2 })*\sin n x * dx\]

Answer 6

\[\int\limits_{0}^{\pi}(1-\frac{ \cos2x }{ 2 }-\frac{ 1 }{ 2 }) *\sin nx * dx\]

Answer 7

solve it further and use sin C & D formula

Answer 8

whats the partial diffy q it came from?

Answer 9

.

Answer 10

@AakashSudhakar

this solves in another 3 or 4 lines:

\[\int\limits_{0}^{π} (\frac{e^{ix} - e^{-ix}}{2i})^{2}(\frac{e^{inx} - e^{-inx}}{2i}) dx\]