Question

integrate sec inverse x​

Answer 1

Given Integrand,

$\displaystyle \sf \int {sec}^{ - 1} x \: dx$

Integrating by parts,

$\longrightarrow \displaystyle \sf {sec}^{ - 1} x \int dx - \int \bigg( \dfrac{d( {sec}^{ - 1}x) }{dx} \int dx \bigg) dx \\ \\ \longrightarrow \displaystyle \sf x {sec}^{ - 1} x - \int \dfrac{x}{ |x| \sqrt{ {x}^{2} - 1}} dx \\ \\ \longrightarrow \displaystyle \sf x {sec}^{ - 1} x - \int \frac{dx}{ \sqrt{ {x}^{2} - 1}} \\ \\ \longrightarrow \displaystyle \boxed{ \boxed{\sf x {sec}^{ - 1} x - ln(x + \sqrt{ {x}^{2} } + 1) + c}}$

Formula Used :

Let x be an independent function and u & v be dependent function which are to be integrated. In accordance with ILATE,

$\bullet \: \boxed{ \boxed{ \displaystyle \sf \int uv \: dx = u \int v \: dx - \int \bigg( \dfrac{du}{dx} \int vdx \bigg)dx \: }}$