Question

Find the value of the following integral.

$\int_0^{\frac{\pi }{4}} {\dfrac{{\sec x}}{{{{\left( {\sec x + \tan x} \right)}^2}}}dx} $

$\left( a \right)1 + \sqrt 2 $

$\left( b \right) - 11 + \sqrt 2 $

$\left( c \right)$-2

$\left( d \right)$ None of these.

Answer 1

Hint: In this particular question first simplify the integral by using some of the basic trigonometric properties such as, tan x = (sin x/cos x) and sec x = (1/cos x), and use the concept of integration by substitution so use these concepts to reach the solution of the question.

Complete step by step solution:

Given integral,

$\int_0^{\frac{\pi }{4}} {\dfrac{{\sec x}}{{{{\left( {\sec x + \tan x} \right)}^2}}}dx} $

Let,

$I = \int_0^{\frac{\pi }{4}} {\dfrac{{\sec x}}{{{{\left( {\sec x + \tan x} \right)}^2}}}dx} $

Now take sec x common from the denominator of the above integral we have,

$ \Rightarrow I = \int_0^{\frac{\pi }{4}} {\dfrac{{\sec x}}{{{{\sec }^2}x{{\left( {1 + \dfrac{{\tan x}}{{\sec x}}} \right)}^2}}}dx} $

Now simplify the above integral using property that, tan x = (sin x/cos x) and sec x = (1/cos x) so we have,

$ \Rightarrow I = \int_0^{\frac{\pi }{4}} {\dfrac{1}{{\sec x{{\left( {1 + \dfrac{{\dfrac{{\sin x}}{{\cos x}}}}{{\dfrac{1}{{\cos x}}}}} \right)}^2}}}dx} $

$ \Rightarrow I = \int_0^{\frac{\pi }{4}} {\dfrac{{\cos x}}{{{{\left( {1 + \sin x} \right)}^2}}}dx} $................. (1)

Now, let (1 + sin x) = t............... (2)

Now differentiate equation (1) w.r.t x we have,

Therefore, $\dfrac{d}{{dx}}\left( {1 + \sin x} \right) = \dfrac{{dt}}{{dx}}$

Now as we know that the differentiation of constant is zero and differentiation of sin x is cos x w.r.t x, so use this property in the above equation we have,

Therefore, cos x dx = dt................. (3)

Now change the integration limits we have,

When, x = 0

So from equation (2) we have,

$ \Rightarrow 1 + \sin 0 = t$

$ \Rightarrow t = 1,\left[ {\because \sin 0 = 0} \right]$

When, x = $\dfrac{\pi }{4}$

So from equation (3) we have,

$ \Rightarrow 1 + \sin \dfrac{\pi }{4} = t$

$ \Rightarrow t = 1 + \dfrac{1}{{\sqrt 2 }},\left[ {\because \sin \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}} \right]$

So integration limits is from $(1 \text{ to }(1 + \dfrac{1}{{\sqrt 2 }}))$

Now substitute the values from equation (2) and (3) in equation (1) and also changed the integration limits we have,

$ \Rightarrow I = \int_1^{1 + \frac{1}{{\sqrt 2 }}} {\dfrac{1}{{{{\left( t \right)}^2}}}dt} $

$ \Rightarrow I = \int_1^{1 + \frac{1}{{\sqrt 2 }}} {{t^{ - 2}}dt} $

Now integrate it using the property $\int {{t^n}dt = \dfrac{{{t^{n + 1}}}}{{n + 1}} + C} $, where c is some integration constant, so use this property in the above equation we have,

$ \Rightarrow I = \left[ {\dfrac{{{t^{ - 2 + 1}}}}{{ - 2 + 1}}} \right]_1^{1 + \frac{1}{{\sqrt 2 }}}$

$ \Rightarrow I = \left[ {\dfrac{{{t^{ - 1}}}}{{ - 1}}} \right]_1^{1 + \frac{1}{{\sqrt 2 }}}$

$ \Rightarrow I = - \left[ {\dfrac{1}{t}} \right]_1^{1 + \frac{1}{{\sqrt 2 }}}$

Now apply integration limits we have,

$ \Rightarrow I = - \left[ {\dfrac{1}{{1 + \dfrac{1}{{\sqrt 2 }}}} - 1} \right]$

$ \Rightarrow I = - \left[ {\dfrac{{1 - 1 - \dfrac{1}{{\sqrt 2 }}}}{{1 + \dfrac{1}{{\sqrt 2 }}}}} \right] = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{1 + \dfrac{1}{{\sqrt 2 }}}} = \dfrac{1}{{1 + \sqrt 2 }}$

Now multiply and divide by ($1 - \sqrt 2 $) we have,

$ \Rightarrow I = \dfrac{1}{{1 + \sqrt 2 }} \times \dfrac{{1 - \sqrt 2 }}{{1 - \sqrt 2 }}$

Now in denominator use the property that, $\left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2}$

$ \Rightarrow I = \dfrac{{1 - \sqrt 2 }}{{{1^2} - {{\left( {\sqrt 2 } \right)}^2}}} = \dfrac{{1 - \sqrt 2 }}{{1 - 2}} = \dfrac{{1 - \sqrt 2 }}{{ - 1}} = - 1 + \sqrt 2 $

This is the required answer.

Hence option (D) none of these is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that choose the substitution term wisely so that the integral convert into the standard integral so that we can solve this easily by using standard integral formula i.e. $\int {{t^n}dt = \dfrac{{{t^{n + 1}}}}{{n + 1}} + C} $, always remember to change the integral limits as above.