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Question

Evaluate secx{sin(2x+α)+sinα}dx

A
2[(tanx+tanα)secα]1/2+C
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B
2[(tanxtanα)cotα]1/2
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C
2[(tanxtanα)secα]1/2
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D
2[(tanx+tanα)secα]1/2
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Solution

The correct option is D 2[(tanx+tanα)secα]1/2+C
sin(2x+α)+sinα=2sin(x+α)cosx=2cosx[sinxcosα+cosxsinα]

=2cos2x[tanxcosα+sinα]

I=secxdx2cosx[tanxcosα+sinα]1/2 =12sec2xdx[tanxcosα+sinα]1/2

Put tanxcosα+sinα=t

sec2xcosαdx=dt I=12cosαdtt=12cosα2t=2[tanxcosα+sinαcos2α]1/2 =2[tanxsecα+secαtanα]1/2+c =2[(tanx+tanα)secα]1/2+c

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