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I (n) = ∫(sin)^n (x) dx in terms of I (n-2)

Let u = sinn-1 x

du = (n-1)sinn-2 x cos x dx

v = - cos x

dv = sinx dx

so integration by parts give:

∫〖sin〗^n x dx= -cos〖x 〖sin〗^(n-1) x+(n-1) ∫〖sin〗^(n-2) 〗 x cos^2〖x dx〗

Since cos2x = 1 – sin2x, we have:

∫〖sin〗^n x dx= -cos〖x 〖sin〗^(n-1) x+(n-1) ∫〖sin〗^(n-2) 〗 x dx-(n-1)∫〖sin〗^n x dx

We solve this equation by taking the last term on the right side to the left side:

∫〖sin〗^n x dx= -cos〖x 〖sin〗^(n-1) x+(n-1)∫〖〖sin〗^(n-2) x dx〗〗

Is this right?