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.

1 Answer
Guillaume L.
Jun 6, 2018

#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 !

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