Question

Evaluate the following integral
$\int_0^{\dfrac{\pi }{4}} {x.{{\sec }^2}xdx} $
$\left( a \right)\dfrac{\pi }{4} + \log \sqrt 2 $
$\left( b \right)\dfrac{\pi }{4} - \log \sqrt 2 $
$\left( c \right)1 + \log \sqrt 2 $
$\left( d \right)1 - \dfrac{1}{2}\log 2$

Answer 1

Hint: In this particular type of question use the concept that, during the integration choose first and second function wisely so that we can apply directly standard integration and differentiation formula than apply integration by parts formula i.e. $\int {{1^{st}}{{.2}^{nd}}dx} = {2^{nd}}\int {{1^{st}}dx} - \int {\left( {\left( {\dfrac{d}{{dx}}{2^{nd}}} \right)\int {{1^{st}}dx} } \right)dx} + C$ so use these concepts to reach the solution of the question.

Complete step by step answer:
Let,
I = $\int_0^{\dfrac{\pi }{4}} {x.{{\sec }^2}xdx} $
Now integrate the above equation by integration by parts we have,
As we know that the formula for integration by parts is given as,
$ \Rightarrow \int {{1^{st}}{{.2}^{nd}}dx} = {2^{nd}}\int {{1^{st}}dx} - \int {\left( {\left( {\dfrac{d}{{dx}}{2^{nd}}} \right)\int {{1^{st}}dx} } \right)dx} + C$, where C is some integration constant.
Now in the above equation (${\sec ^2}x$) is the ${1^{st}}$ function and (x) is the ${2^{nd}}$ function so we have,
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{4}} {x.{{\sec }^2}xdx} = \left[ {x\int {{{\sec }^2}xdx} } \right]_0^{\dfrac{\pi }{4}} - \int\limits_0^{\dfrac{\pi }{4}} {\left( {\left( {\dfrac{d}{{dx}}x} \right)\int {{{\sec }^2}xdx} } \right)dx} \]
Now as we know that the integration of ${\sec ^2}x$ is tan x and differentiation of x w.r.t. x is 1, so use this properties in the above equation we have,
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{4}} {x.{{\sec }^2}xdx} = \left[ {x\tan x} \right]_0^{\dfrac{\pi }{4}} - \int\limits_0^{\dfrac{\pi }{4}} {\left( {\left( 1 \right)\tan x} \right)dx} \]
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{4}} {x.{{\sec }^2}xdx} = \left[ {x\tan x} \right]_0^{\dfrac{\pi }{4}} - \int\limits_0^{\dfrac{\pi }{4}} {\tan xdx} \]
Now as we know that the integration of tan x is (-log (cos x)), so use this properties in the above equation we have,
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{4}} {x.{{\sec }^2}xdx} = \left[ {x\tan x} \right]_0^{\dfrac{\pi }{4}} - \left[ { - \log \cos x} \right]_0^{\dfrac{\pi }{4}}\]
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{4}} {x.{{\sec }^2}xdx} = \left[ {x\tan x} \right]_0^{\dfrac{\pi }{4}} + \left[ {\log \cos x} \right]_0^{\dfrac{\pi }{4}}\]
Now apply the integration limits we have,
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{4}} {x.{{\sec }^2}xdx} = \left[ {\dfrac{\pi }{4}\tan \dfrac{\pi }{4} - 0} \right] + \left[ {\log \cos \dfrac{\pi }{4} - \log \cos 0} \right]\]
Now simplify this we have,
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{4}} {x.{{\sec }^2}xdx} = \dfrac{\pi }{4} + \left[ {\log \left( {\dfrac{1}{{\sqrt 2 }}} \right) - \log \left( 1 \right)} \right]\], $\left[ {\because \tan \dfrac{\pi }{4} = 1,\cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }},\cos 0 = 1} \right]$
Now as we know that the value of log 1 = 0 and use the property that log (1/x) = -log x, so use this properties in the above equation we have,
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{4}} {x.{{\sec }^2}xdx} = \dfrac{\pi }{4} + \left[ { - \log \sqrt 2 - 0} \right]\]
\[ \Rightarrow I = \int_0^{\dfrac{\pi }{4}} {x.{{\sec }^2}xdx} = \dfrac{\pi }{4} - \log \sqrt 2 \]
So this is the required answer.
Hence option (B) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always remember basic integration and basic differentiation formula which is used in that question and which is all stated above so first simplify the integration and then integrate it by parts as above then apply these formulas and simplify as above we will get the required answer.