Plz solve this sum

Asked by  | 9th Feb, 2010, 06:18: AM

Expert Answer:

(tannx + tann-2x)dx, and the limits are 0 to π/4

tannx + tann-2x = tannx(1 + 1/tan2x) = tannx((tan2x + 1)/tan2x) = tann-2(sec2x)

(tannx + tann-2x)dx = tann-2(sec2x) dx 

Put tanx = y

sec2xdx = dy, and the limits x = 0, y = 0 and x = π/4, y = 1.

yn-2dy = yn-1/(n-1) + c

Put the limits,

yn-2dy = 1/(n-1), where n is an integer except 1, i.e. n N - {1}

Regards,

Team,

TopperLearning.

 

 

Answered by  | 9th Feb, 2010, 10:14: AM

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