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Question

Evaluate ex(1+sinx1+cosx)dx

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Solution

Consider, I=ex(1+sinx1+cosx)dx

=ex⎜ ⎜1+2sinx2.cosx22cos2x2⎟ ⎟dx

=ex(12sec2x2+tanx2)dx

=extanx2dx+ex2sec2x2dx

=extanx2ex2sec2x2dx+ex2sec2x2dx+c [c being integrating constant] [Using by-parts method]

=extanx2+c.

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