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

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william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

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

mikeadam1234's profile pic

mikeadam1234 | Student, Undergraduate | (Level 2) eNoter

Posted on

Ok. Gotcha, that's why we had had the n not equal to -1, else n+1 would be zero.

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neela | High School Teacher | (Level 3) Valedictorian

Posted on

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

 

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