3 Answers | Add Yours
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
Ok. Gotcha, that's why we had had the n not equal to -1, else n+1 would be zero.
To integrate (cosx)^n*sinx.
Let I = Int(cosx)^n sinx dx
We put cosx = t. Then by differentiating both sides,we get:
-sinxdx = dt.
sinxdx = -dt
So I = Int (t^n) *-dt
I = -(1/(n+1))t^(n+1).
I = (1/(n+1)(cosx)^(n+1) + Const
We’ve answered 319,828 questions. We can answer yours, too.Ask a question