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Question

Evaluate: π20sinx.cosxsin4x+cos4xdx

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Solution

Let I=π/20sinxcosxsin4x+cos4xdx

I=π/20sinxcosxcos4xsin4xcos4x+cos4xcos4xdx

=π/20sinxcos3xtan4x+1dx

=π/20tanxsec2xtan4x+1dx

Put tan2x=u, then, 2tanxsec2xdx=du.

I=12π/20duu2+1

=12[tan1u]π/20

=12[tan1(tan2x)]π/20

=12[tan1(tan2π2)tan1(tan20)]

=12[tan1()tan1(0)]

=12[π20]

=π4


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