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.