Question

Evaluate \[\int\limits_0^x {\dfrac{{x\,\,\tan \,x}}{{\sec x\, + \,\tan \,x}}dx} \]

Answer 1

Hint:
\[\left( {tan{\text{ }}x} \right)\] can be written in terms of \[\left( {sin{\text{ }}x} \right)\]and \[\left( {cos{\text{ }}x} \right)\] . Also try to solve the question by breaking it into simple steps like one can rationalize the denominator and write the answer after performing certain mathematical calculations.

Complete step by step solution:
Given:
\[\int\limits_0^x {\dfrac{{x\,\,\tan \,x}}{{\sec x\, + \,\tan \,x}}dx} \]
Stepwise solution:
Let us rationalize the given equation first. So that our question becomes much easier
$\int\limits_0^x {\dfrac{{x\,\tan x\,(\sec \,x\, - \,\tan x)}}{{(\sec x\, - \,\tan \,x)\,(\sec \,x + \tan \,x)}}dx} $
$ \Rightarrow \,\,\,\,\int\limits_0^x {\dfrac{{x\,\tan x\sec x - \,x\,{{\tan }^2}x)}}{{{{\sec }^2}x - {{\tan }^{2x}}}}dx} $
\[ \Rightarrow \,\,\,\,\int\limits_0^x {\dfrac{{x\,\tan x.\sec \,x - x\,{{\tan }^2}x}}{1}dx} \] $\left[ {\because \,\,{{\sec }^2}x + {{\tan }^2}x = 1} \right]$
$ \Rightarrow \,\,\,\,\int\limits_0^x {\left( {x\tan x.\,\sec x - x({{\sec }^2}x - 1).dx} \right)} $ [Since, ${\tan ^2}x = {\sec ^2}x - 1$ ]
\[ \Rightarrow \,\,\int\limits_0^x {x\tan x.\sec x\,dx} \, - \,\int\limits_0^x {x{{\sec }^2}x\,dx} \, - \,\,\int\limits_0^x {x.dx} \]
Now, using the [I. L. A. T. E] rule we will get the final answer.
The [I. L. A. T. E] rule is \[u\,\int {v\,dx} \, - \,\int {\left( {\dfrac{{du}}{{dx}}\,.\,\int {v\,dx} \,dx} \right)} \]
Thus, on applying it we get,
\[x\,\int\limits_0^x {\tan x\,\sec x\,dx} - \,\int\limits_0^x {\left( {\dfrac{d}{{dx}}x\,\int\limits_0^x {\tan x.\sec x\,dx} } \right)} \, - \,\,\left[ {x\,\int\limits_0^x {{{\sec }^2}x\,dx - \,\int {\left( {\dfrac{d}{{dx}}x.\,\,\int\limits_0^x {{{\sec }^2}x\,dx} } \right)} } \,dx} \right]\, + \,\int\limits_0^x {x\,dx} \]
\[ = x\sec x\, - \,\ln |\sec x + \tan x|\, - \,\left[ {x\tan x - \ln |\sec x|} \right]\, + \,\dfrac{{{x^2}}}{2}\]
$ = x\sec x - x\tan x - \,\ln (\sec x + \tan x) + \ln \,|\sec x| + \dfrac{{{x^2}}}{2}$
$ = \,x(\sec x - \tan x)\, - \,\ln (\sec x(\sec x + \tan x)) + \dfrac{{{x^2}}}{2} + C$
Since, $\ln \,M + \ln \,N = \,\ln (M.V)$
Therefore, the answer is
$ \Rightarrow x(\sec x - \tan x) - \ln |\sec x(\sec x + \tan x)| + \,\dfrac{{{x^2}}}{2} + C$

Note: In this type of question student often gets confused regarding [I. L. A. T. E] rule and they make a mistake by not rationalizing the equation at initial stage. Do not repeat the mistake and do not forget to write the constant “C” after the answer as without it your answer will be incorrect. Also remember that ln (M+ N) is never equal to ln multiplied by ln N. This is the most occurring conception among students. So, use these logarithmic statements carefully.