Question

Find the value of $\int {{{\tan }^{ - 1}}\left( {\sec x + \tan x} \right)dx}$.

Answer 1

Hint: Convert $\left( {\sec x + \tan x} \right)$ in terms of $\tan x$ and then integrate.

Let

$$I = \int {{{\tan }^{ - 1}}\left( {\sec x + \tan x} \right)dx} \\\ \\\ I = \int {\tan ^{ - 1}}\left( {\dfrac{1}{{\cos x}} + \dfrac{{\sin x}}{{\cos x}}} \right)dx \\\ \\\ \\\ I = \int {{{\tan }^{ - 1}}\left( {\dfrac{{1 + \sin x}}{{\cos x}}} \right)dx} \\\$$

Converting into their half-angles then we have
$I = \int {{{\tan }^{ - 1}}\left( {\dfrac{{{{\sin }^2}\dfrac{x}{2} + {{\cos }^2}\dfrac{x}{2} + 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}{{{{\cos }^2}\dfrac{x}{2} - {{\sin }^2}\dfrac{x}{2}}}} \right)} dx \\\ \\\ I = \int {{{\tan }^{ - 1}}\left( {\dfrac{{{{\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)}^2}}}{{\left( {\cos \dfrac{x}{2} - \sin \dfrac{x}{2}} \right)\left( {\cos \dfrac{x}{2} + \sin \dfrac{x}{2}} \right)}}} \right)} \\\$
Cancelling common terms in numerator and denominator, we have
$I = \int {{{\tan }^{ - 1}}} \left( {\dfrac{{\cos \dfrac{x}{2} + \sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2} - \sin \dfrac{x}{2}}}} \right)dx$

$I = \int {{{\tan }^{ - 1}}\left( {\dfrac{{1 + \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}}}{{1 - \dfrac{{\sin \dfrac{x}{2}}}{{\cos \dfrac{x}{2}}}}}} \right)} dx$
$I = \int {{{\tan }^{ - 1}}\left( {\dfrac{{1 + \tan \dfrac{x}{2}}}{{1 - \tan \dfrac{x}{2}}}} \right)} dx$

Since $\dfrac{{1 + \tan \dfrac{x}{2}}}{{1 - \tan \dfrac{x}{2}}} = \tan \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)$ we have

$$I = \int {{{\tan }^{ - 1}}\left( {\tan \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)} \right)dx} \\\ \\\ I = \int {\left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)dx} \\\ \\\ I = \dfrac{\pi }{4}x + \dfrac{1}{4}{x^2} + c \\\$$

Thus, $\int {{{\tan }^{ - 1}}\left( {\sec x + \tan x} \right)dx} = \dfrac{\pi }{4}x + \dfrac{1}{4}{x^2} + c$.

Note: Do not forget to add integration constant after doing integration.