What is the integration of sec inverse root x ?

Solution :

We have, I = \(sec^{-1}\sqrt{x}\) dx

Let x = \(sec^2t\)

dx = \(2sec^2 t tan t\) dt

I = t.\(2sec^2 t tan t\) dt

u = t  and v = \(tan^2 t\)

I = \(\int\) u.dv = u.v – \(\int\) v.du = \(t.tan^2 t\) – \(\int\) \(tan^2 t\) dt

I = \(t.tan^2 t\) – \(\int\) \(sec^2 t – 1\) dt

I = \(t.tan^2 t\) – \(\int\) \(sec^2 t\) + \(int\) 1 dt

I = \(t.tan^2 t\) – tan t + t + C = \(t.sec^2 t\) – tan t + C

I = \(xsec^{-1}\sqrt{x}\) – \(\sqrt{sec^2t – 1}\) + C

I = \(xsec^{-1}\sqrt{x}\) – \(\sqrt{x – 1}\) + C


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