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Question

Prove the following question.

π20sin3 x dx=23.


Solution

Let I = π20sin3 x dx=π20sin2 x. sin x dx   =π20(1cos2x)sin x dx    ( sin2x=1cos2x)Put,cos x=t  sin x dx=dtWhen x=0 t=cos 0=1,when x=π2=0  I=π20(1cos2x)sin x dx=01(1t2)(dt)=[tt33]01={(00)(113)}=23.   Hence proved.


Mathematics

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