1.

Obtain the reduction formula for `intcos^(n)xdx`

Answer» Let `I_(n) = intcos^(n)xdx`
`I_(n)=intcosx(cosx)^(n-1)dx`
`I_(n)=(sinx)(cosx)^(n-1)-int(n-1)(cosx)^(n-2)(-sinx)sinxdx`
`I_(n)=(sinx)(cosx)^9n-1+(n-1)int(cosx)^(n-2)dx-(n-1)int(cosx)^(n)dx`
`I_(n) = (sinx)(cosx)^(n-1)+(-1)I_(n-2)-(n-1)I_(n-2)`
`I_(n)+(n-1)I_(n)=(sinx)(cosx)^(n-1)+(n-1)I-(n-2)`
`I_(n)=((sinx)cosx))^(n-1)+(n-1)I_(n-2)`
`I_(n) = ((sinx)(cosx)^(n-1))/(n) + (n-1)/(n)I-(n-2), nge2`


Discussion

No Comment Found

Related InterviewSolutions