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₀âˆ«^pi/4(e^tanx(secx-sinx))dx

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Manish Shankar ·2008-11-02 00:02:38

âˆ«e^tanxsecxdx-âˆ«e^tanxsinxdx

solving the second using âˆ«e^tanx as first function

âˆ«e^tanxsinx=e^tanxâˆ«sinxdx-âˆ«[(de^tanx/dx)âˆ«sinxdx]dx

=-e^tanxcosx-[-âˆ«e^tanxsec²xcosxdx

=-e^tanxcosx+âˆ«e^tanxsecxdx

âˆ«e^tanxsecx-âˆ«e^tanxsinx=âˆ«e^tanxsecx-(-e^tanxcosx+âˆ«e^tanxsecxdx)

=[e^tanxcosx]₀^Ï€/4=e/âˆš2 - 1

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Abhishek Priyam ·2008-11-02 00:07:58

Gr8,
options were e/âˆš2 - 1 , e/âˆš2+1, 0 and none
so i drew graph of given fn and eliminated 0, and e/âˆš2+ 1 as area was seeming much less.... so i choosed: e/âˆš2 - 1

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Basic Integral Calculus Operators Symbols Matrix Cases

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∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

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