- #1
natashajane
- 7
- 0
Use integration by parts to express:
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?
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?