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.
I have come this far
Then we substitute,
I can't figure out how to compute rest of it.
1 Answer
Jun 6, 2018
Explanation:
Let
\0/ here's our answer !