Question

integration of x^2 sec invers x​

Answer 1

Given Integrand,

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

According to ILATE,

arcsec(x) is the first expression and x is the second one.

Integrating by parts,

$\implies \sf \: \displaystyle \sf {sec}^{ - 1} (x) \int {x}^{} dx - \int \bigg( \dfrac{d( {sec}^{ - 1}x) }{dx} \int {x}^{} dx\bigg)dx \\ \\ \implies \sf \: \displaystyle \sf \dfrac{1}{2} {x}^{2} {sec}^{ - 1} (x) - \dfrac{1}{2} \int \dfrac{ {x}^{2} }{ |x| \sqrt{ {x}^{2} - 1 } }dx \\ \\ \implies \sf \: \displaystyle \sf \dfrac{1}{2} {x}^{2} {sec}^{ - 1} (x) - \dfrac{1}{2} \int \dfrac{ {x}^{} }{ \sqrt{ {x}^{2} - 1 } } dx$

Consider,

$\displaystyle \sf \: l = \int \dfrac{ {x}^{} }{ \sqrt{ {x}^{2} - 1 } }$

Let t = x² - 1.

Differentiating both sides w..r.t x,

$\longrightarrow \sf \: dt = 2xdx \\ \\ \longrightarrow \sf \: \dfrac{dt}{2x} = dx$

Further,

$\displaystyle \sf \: l = \int \dfrac{ {x}^{} }{ \sqrt{ {x}^{2} - 1 } } \\ \\ \longrightarrow \displaystyle \sf \: l = \int \dfrac{x}{ \sqrt{t} } \dfrac{dt}{2x} \\ \\ \longrightarrow \displaystyle \sf \: l = \dfrac{1}{2} \int {t}^{ {}^{ - \frac{1}{2} } }dt \\ \\ \longrightarrow \displaystyle \sf \: l = \sqrt{x {}^{2} - 1}$

Thus,

$\boxed{ \boxed{ \displaystyle \sf \int {x}^{} {sec}^{ - 1} (x)dx = \dfrac{ {x}^{2} {sec}^{ - 1}(x) }{2} - \dfrac{1}{2}( \sqrt{ {x}^{2} - 1} ) + c}}$

..

Answer 2

I hope it helps all.

Thanks