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Kindly consider the following

∫e−3xcos3xdx

0 1 View | Posted 3 months ago
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    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

    This is a Long Answer Type Questions as classified in NCERT Exemplar

    Sol:

    Let????I=?e?3xIIcos3xIdx????????????????=cos3x.?e?3x?dx??(D(cos3x).?e?3x?dx)dx????????????????=cos3x.e?3x?3??(3cos2x(?sinx).e?3x?3)dx????????????????=?13e?3xcos3x??cos2xsinx.e?3xdx????????????????=?13e?3xcos3x??(1?sin2x)sinx.e?3xdx????????????????=?13e?3xcos3x??sinx.e?3xdx+?sin3xI.e?3xIIdx????????????????=?13e?3xcos3x??sinx.e?3xdx+sin3x?e?3xdx??(D(sin3x).?e?3x?dx)dx????????????????=?13e?3xcos3x??sinx.e?3xdx+sin3x.e?3x?3??(3sin2x.cosx.e?3x?3)dx????????????????=?13e?3xcos3x??sinx.e?3xdx?13e?3xsin3x+?sin2xcosx.e?3xdx????????????????=?13e?3xcos3x??sinx.e?3xdx?13e?3xsin3x+?(1?cos2x)cosx.e?3xdx?????????????I=?13e?3xcos3x?[sinx.e?3x?3??cosx.e?3x?3dx]?13e?3xsin3x+?cosx.e?3xdx??cos3x.e?3xdx????????????????=?13e?3xcos3x+sinx.e?3x3??cosx.e?3x?3dx?13e?3xsin3x+?cosx.e?3xdx?I?????????2I=e?3x?3[cos3x+sin3x]?[sinx.e?3x3??cosx.e?3x?3dx]+?cosx.e?3xdx????????????????=e?3x?3[cos3x+sin3x]+13sinx.e?3x?13?cosx.e?3xdx+?cosx.e?3xdx????????2I=e?3x?3[cos3x+sin3x]+13sinx.e?3x+23?cosx.e?3xdxNow,?I1=23?cosxI.e?3xIIdx????????????????=23[cosx.?e?3xdx??(D(cosx).?e?3x?dx)dx]????????????????=23[cosx.e?3x?3???sinx.e?3x?3dx]????????????????=23[cosx.e?3x?3?13?sinx.e?3xdx]

    =2?9cosx.e?3x?29?sinx.e?3xdx???????????I1=2?9cosx.e?3x?29[sinx.e?3x?3??cosx.e?3x?3dx]???????????I1=?29cosx.e?3x+227sinx.e?3x?227?cosx.e?3xdx???????????I1=?29cosx.e?3x+227sinx.e?3x?19.23?cosx.e?3xdx???????????I1=?29cosx.e?3x+227sinx.e?3x?19.I1I1+19I1=?29cosx.e?3x+227sinx.e?3x???????10I19=?29cosx.e?3x+227sinx.e?3x???????????I1=?110cosx.e?3x+115sinx.e?3xSo,?????2I=?13e?3x[sin3x+cos3x]+13sinx.e?3x?110cosx.e?3x+115sinx.e?3x?????????????I=?16e?3x[sin3x+cos3x]+16sinx.e?3x?120cosx.e?3x+130sinx.e?3x????????????????=?16e?3x[sin3x+cos3x]+15sinx.e?3x?120cosx.e?3x????????????????=e?3x24[sin3x?cos3x]+3e?3x40[sinx

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= ∑? n * e? [-e? - (-e¹? )]
= ∑? n * e? (e¹? - e? )
= ∑? n * e? * e? (e - 1)
= (e - 1) ∑? n
= (e - 1) * (0 + 1 + 2 + . + 9)
= (e - 1) * (9 * 10 / 2)
= 45 (e - 1)

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