Question

Need to integrate this with substitution `int_(pi/2)^pi6sinx(cosx+1)dx`?

I have come this far

`u = cosx +1`
`du = -sinxdx`

Then we substitute,

`-6int_(pi/2)^piu^5 du`

`-6[u^6/6]_(pi/2)^pi`

I can't figure out how to compute rest of it.

Answer 1

`int_(pi/2)^pi6sin(x)(cos(x)+1)dx=3`

Explanation:

`int_(pi/2)^pi6sin(x)(cos(x)+1)dx`
`=6int_(pi/2)^pi(sin(x)cos(x)+sin(x))dx`
`=6int_(pi/2)^pisin(x)cos(x)dx+6int_(pi/2)^pisin(x)dx`
Let `u=sin(x)`
`du=cos(x)dx`
`=6intudu +[-6cos(x)]_(pi/2)^pi`

`=[6(sin(x)²)/2-6cos(x)]_(pi/2)^pi`
`=6*(0)/2-6*(-1)-(6*(1²)/2-6*0)`
`=6-3`
`=3`
\0/ here's our answer !