# 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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