How do we find the integral of (cos x)^n sin x for n not equal to -1 ?

We need to find Int [ (cos x)^n sin x
dx]

First let us put u = cos x
,

we get du/dx  = -sin x

or du
= -sin x dx

Now in Int [(cos x)^n sin x
dx]

replace cos x = u and sin x dx = -
du

=> - Int [ (cos x)^n du
]

=> - Int [ u^n du
]

=> - u^(n+1) / (n+1) +
C

=> - [(cos x)^(n+1)] / (n+1) +
C

Therefore Int [ (cos x)^n sin x dx] = -
[(cos x)^(n+1)] / (n+1) + C