Question

Evaluate the integral\(\int {{{\tan }^2}} x\sec xdx\)

Short answer

The answer is \(\frac{{\sec x\tan x}}{2} - \frac{{\ln |\tan x + \sec x|}}{2} + c\) , where c is an Integral Constant.

First, substitute in\(\sec x\) with u and also put \(\tan x\) in terms of \(\sec x\) whenever necessary and solve the Integral.

Step-by-step solution

Putting \(\sec x = u\) 

\(\begin{aligned}{l} \Rightarrow \sec x\tan xdx &= du\\ \Rightarrow \tan x\sec xdx &= du \to 1\end{aligned}\)

So,

\(\begin{aligned}{l}\int {{{\tan }^2}} x\sec xdx\\ &= \int {\tan x.\tan x.\sec xdx} \\ &= \tan x.dx\end{aligned}\)

Converting \(\tan x\)in \(\sec x\)using trigonometric identity and putting \(\sec x = u\)

\(\begin{aligned}{l} &= \int {\tan xdx} \\ &= \int {\left){\vphantom{1{\left( {{{\sec }^2}x - 1} \right)}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{{\left( {{{\sec }^2}x - 1} \right)}}}} .dx\\ &= \int {\left){\vphantom{1{\left( {{u^2} - 1} \right)}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{{\left( {{u^2} - 1} \right)}}}} .dx\end{aligned}\)

Solving the integral using formula

\(\begin{aligned}{l} &= \int {\left){\vphantom{1{\left( {{u^2} - 1} \right)}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{{\left( {{u^2} - 1} \right)}}}} .du\\ &= \frac{1}{2}u\left){\vphantom{1{{u^2} - 1}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{{{u^2} - 1}}} - \frac{1}{2}\ln |u + \left){\vphantom{1{{u^2} - 1}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{{{u^2} - 1}}}| + c\end{aligned}\)

Putting back \(\sec x = u\)

\(\)\(\begin{aligned}{l} &= \frac{1}{2}\sec x\left){\vphantom{1{\left( {{{\sec }^2}x - 1} \right)}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{{\left( {{{\sec }^2}x - 1} \right)}}} - \frac{1}{2}\ln |\sec x + \left){\vphantom{1{\left( {{{\sec }^2}x - 1} \right)}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{{\left( {{{\sec }^2}x - 1} \right)}}}| + c\\ &= \frac{1}{2}\sec x\left){\vphantom{1{\left( {{{\sec }^2}x - 1} \right)}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{{\left( {{{\sec }^2}x - 1} \right)}}} - \frac{1}{2}\ln |\sec x + \tan x| + c\end{aligned}\)

Hence,\(\int {{{\tan }^2}} x\sec xdx\)

=\(\frac{1}{2}\sec x\left){\vphantom{1{\left( {{{\sec }^2}x - 1} \right)}}}\right.

\!\!\!\!\overline{\,\,\,\vphantom 1{{\left( {{{\sec }^2}x - 1} \right)}}} - \frac{1}{2}\ln |\sec x + \tan x| + c\)