intigrate w.r.t x

Asked by sumahr | 23rd Dec, 2009, 07:00: PM

Expert Answer:

Put tan-1(2x/(1 - x2)) = y

dy/dx = (1/(1 + (2x/(1 - x2))2) [(1 - x2).2 - 2x (-2x)]/(1 - x2)2

{Note the use of chain rule and standard result d(tan-1x)/dx = 1/(1 + x2)}

Simplifying gives,

dy/dx = 2/(1+x2)...(1)

2x/(1-x2) = tan y

tan y x2 +2x - tan y = 0

x = (-1±sec y)/tan y.

1 + x2 = 2sec y(sec y ± 1)/tan2 y = 2(cosec2y ± cosec2y cos y)

2(cosec2y ± cosec2y cos y) dy = dx.

tan-1(2x/(1 - x2)) dx = 2y(cosec2y ± cosec2y cos y) dy

This can be separated and integrated by parts.

Regards,

Team,

TopperLearning.

Answered by  | 23rd Dec, 2009, 11:30: PM

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