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Question

Solve π20cos2x dxcos2x+4sin2x

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Solution

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π2ocos2xdxcos2x+4sin2x

=π2ocos2xdxcos2x+4(1cos2x)

=π2ocos2xdx3cos2x+4

=13π2o(3cos2x+4)4dx3cos2x+4

=13π2o(14dx43cos2x)

=13[x]π20+43π2o(dx43cos2x)

=π6+43π2o(sec2xdx4sec2x3)

Dividing Nr and Dr by cos2x

=π6+43π2o(sec2xdx4(1+tan2x)3)

put tanx=t

sec2xdx=dt

=π6+43o(dt4(1+t2)3)

=π6+430(dt2t2+1)

=π6+43×12[tan1(2t)]0=π3


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