Question

Evaluate $\int {x\sec x.\tan xdx = }$

$$A.{\text{ xsecx + log}}\left| {\tan (\dfrac{\pi }{2} + \dfrac{x}{2}} \right| + c \\\ B.{\text{ xsecx + log}}\left| {\tan (\dfrac{\pi }{2} + \dfrac{x}{2}} \right| + c \\\ C.{\text{ xsecx + log}}\left| {\tan (\dfrac{\pi }{2} + x} \right| + c \\\ B.{\text{ xsecx + log}}\left| {\tan \dfrac{x}{2}} \right| + c \\\$$

Answer 1

We are aware that the reverse process of differentiation is integration. The problem given can be solved with the help of some basic integration and differentiation. We will use integration by parts method. Further to solve questions where multiple terms are present as product terms inside integration, substitution technique can be used.

Formula used :
The above problem can be solved by integration by parts. The formula of integration by part is
$\int {udv = uv - \int {vdu} }$
Therefore after simplifying

$$\Rightarrow u = x \\\ \Rightarrow du = dx \\\ \Rightarrow dv = \sec x\tan xdx \\\ \Rightarrow v = \sec x \\\$$

By applying this in the given equation we can evaluate it.

Complete step by step solution:
As we know that we need to evaluate $\int {x\sec x.\tan xdx = }$
So it can be written as
$\int {x(\sec x(\tan x))dx}$
Now by using integration by parts

$$\Rightarrow x\int {(\sec x\tan x)dx - \int {1\int {(\sec x\tan x)dx} } } \\\ \Rightarrow x\sec x - \int {\sec xdx} \\\ \Rightarrow x\sec x - \log \left| {\sec x + \tan x} \right| + c \\\ \Rightarrow x\sec x - \log \left| {\tan (\dfrac{\pi }{4} + x} \right| + c \\\$$

Here we will notice an additional term $c$ known as integration constant which we get while performing indefinite integration.
As we know that
$\sec x + \tan x = \tan \left( {\dfrac{\pi }{4} + \dfrac{x}{2}} \right)$
Now applying in in above equation we will get
$x\sec x - \log \left| {\tan \dfrac{\pi }{4} + \dfrac{x}{2}} \right| + c$
Therefore
$\int {x\sec x\tan xdx = x\sec x - \log \left| {\tan (\dfrac{\pi }{2} + \dfrac{x}{2})} \right| + c}$

So, the correct answer is Option B.

Note: While facing such types of problems one should have a good grasp of integration formula. Here we need to recall the definition of tangent in terms of sine and cosine. Most of the problems can be solved by using the right technique easily so do not forget the shorthand methods for multiple applications for integration by parts problems. Here we get the answer as $x\sec x - \ln \left| {\sec x + \tan x} \right| + c$ but keep in mind to simplify and convert the equation into the form it is asked and after simplifying we get $\int {x\sec x\tan xdx = x\sec x - \log \left| {\tan (\dfrac{\pi }{2} + \dfrac{x}{2})} \right| + c}$