solve underroot cotx.

Asked by priyadarshini_selvam | 22nd Mar, 2012, 03:53: PM

Expert Answer:

Take (cot x) = t2 --> -cosec2x dx = 2t dt and we know that cosec2x =1+cot2x=1+t4

so now you have to find integration of [ 2t2/(1+t4) dt ]

= ?[(t^2 + 1) + (t^2 - 1)] / (1 + t^4) dt
= ?(t^2 + 1) / (1 + t^4) dt + ?(t^2 - 1) / (1 + t^4) dt
= ?(1 + 1/t^2) / (t^2 + 1/t^2) dt + ?(1 - 1/t^2) / (t^2 + 1/t^2) dt
= ?(1 + 1/t^2)dt / [(t - 1/t)^2 + 2] + ?(1 - 1/t^2)dt / [(t + 1/t)^2 -2]
Let t - 1/t = u for the first integral => (1 + 1/t^2)dt = du
and t + 1/t = v for the 2nd integral => (1 - 1/t^2)dt = dv
Integral
= ?du/(u^2 + 2) + ?dv/(v^2 - 2)
= (1/?2) arctan (u/?2) + (1/2?2) ln l(v -?2)/(v + ?2)l + c
= (1/?2) arctan [(t^2 - 1)/t?2] + (1/2?2) ln l (t^2 + 1 - t?2) / t^2 + 1 + t?2) + c

We get the answer by sustituting the value of t

Answered by  | 22nd Mar, 2012, 04:21: PM

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