Need to integrate this with substitution int_(pi/2)^pi6sinx(cosx+1)dx∫ππ26sinx(cosx+1)dx?
I have come this far
u = cosx +1u=cosx+1
du = -sinxdxdu=−sinxdx
Then we substitute,
-6int_(pi/2)^piu^5 du−6∫ππ2u5du
-6[u^6/6]_(pi/2)^pi−6[u66]ππ2
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
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