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Question

Find solution in terms of indefinite integration, using substitution
10xcos(tan1x)dx

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Solution

10xcos(tan1x)dx
Let,
tan1x=t
at, x=0,t=0;x=1,t=π4
dt=11+x2dx

10xcos(tan1x)dx

= π40tant×cost×(1+tan2t)dt

= π40tant×cost×(sec2t)dt ......{1+tan2t=sec2t}

= π40sintcos2tdt ......{1cost=sectandtant=sintcost}

= [(cost)(2+1)2+1]π40

=[(cost)11]π40

=[(secπ4)(sec0)]

=2+1

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