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Question

xa3x3dx=

A
23cos1(xa)3/2+C
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B
13sin1(xa)3/2+c
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C
23sin1(xa)3/2+c
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D
13cos1(xa)3/2+c
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Solution

The correct option is B 23sin1(xa)3/2+c
xa3x3dx
1axa1x3a3dx
1a23d(xa)321[(xa)32]2
=23×1a×asin1[(xa)32]+c
=23sin1[(xa)32]+c

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