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Question

Evaluate:
π40esinx(1+tanxsecx)dx

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Solution

given, π40esinx(1+tanxsecx)dx

Let esinxsecx=z

or, dz=(esinx.cosx.secx+esinxsecx.tanx)dx

or, dz=esinx(1+secx.tanx)dx

So the given integration will take the form

π40d(esinxsecx)


=[esinxsecx]π40

=(2e121)

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