Integral Calculus

Reduce sin n ax


I = ∫sin n ax cos m ax dx = ∫sin n-1+1 ax cos m ax dx

I= ∫sin n-1 ax sin ax cos m axdx = ∫sin n-1 ax cos m ax (sin axdx) -------- (1)
Since

d d dx
(cos ax) = - sin ax (ax) = - a sin ax = - a sin ax
dx dx dx
d(cos ax) = - a sin ax dx
-1
d(cos ax) = sin axdx ------ (2)
a
Putting equation (2) in equation (1)

1
I = ∫sin n-1 ax cos m ax ( sin axdx) = - ∫sin n-1 ax cos m ax d(cos ax)
a
-aI = ∫sin n-1 ax cos m ax d(cos ax) ---------- (3)

d d
Since cos m+1 ax = (m + 1)cos m+1-1 ax (cos ax)
dx dx
d
= (m + 1)cos m ax (cos ax)
dx
d d
= (m + 1)cos m ax (cos ax) = (m + 1)cos m ax (cos ax)
dx dx

d d
cos m+1 ax = (m + 1)cos m ax
(cos ax)
dx dx
d cos m+1 ax = (m + 1)cos m ax d(cos ax)
1
d cos m+1 ax = cos m ax d(cos ax) -------------- (4)
(m + 1)