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in part ii i am stuck at this step ..

Use the substitution now

ʃ5sin^3x*cos^2x dx

You have dx and u substitution

= ʃ 5sin^3x(u^2) (du/-sinx)

Cancelling out will give you,

=ʃ-5sin^2x(u^2)

Use trig identity : sin^2x = 1 – cos^2x

=-ʃ5(1 – cos^2x)(u^2)du

=-5ʃ(1 – u^2)(u^2)du

= -5ʃu^2 – u^4 du

=- 5 [(u^3/3) – (u^5/5)]

Put limits in.

-5 {[(1/3 – 1/5) – (0/3 – 0/5)]}

= -5 (2/15)

= -10/15

= -2/3

Area = 2/3 Ans.

*I think I messed up a sign somewhere or its the way it is

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