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Question

Solve 12dxx5x21

A
132(π+7328)
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B
132(π7328)
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C
132(π+732+8)
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D
None of these
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Solution

The correct option is A 132(π+7328)
Let I=1x5x21dx
Put x=sectdx=tantsectdt
I=cos4tdt
Using cosmdx=sin(x)cosm1xm+m1mcos2+xxdx
I=14sintcos3t+34cos2tdx
=14sintcos3t+38(cos2x+12)dx
=38sec1x+14cos(sec1x)sin(sec1x)+38(sin(sec1x)cos(sec1x))
12dxx5x21=132(π+7328)

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