Need to integrate this with substitution int_(pi/2)^pi6sinx(cosx+1)dxππ26sinx(cosx+1)dx?

I have come this far

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

Then we substitute,

-6int_(pi/2)^piu^5 du6ππ2u5du

-6[u^6/6]_(pi/2)^pi6[u66]ππ2

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

1 Answer
Jun 6, 2018

int_(pi/2)^pi6sin(x)(cos(x)+1)dx=3ππ26sin(x)(cos(x)+1)dx=3

Explanation:

int_(pi/2)^pi6sin(x)(cos(x)+1)dxππ26sin(x)(cos(x)+1)dx
=6int_(pi/2)^pi(sin(x)cos(x)+sin(x))dx=6ππ2(sin(x)cos(x)+sin(x))dx
=6int_(pi/2)^pisin(x)cos(x)dx+6int_(pi/2)^pisin(x)dx=6ππ2sin(x)cos(x)dx+6ππ2sin(x)dx
Let u=sin(x)u=sin(x)
du=cos(x)dxdu=cos(x)dx
=6intudu +[-6cos(x)]_(pi/2)^pi=6udu+[6cos(x)]ππ2

=[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 !