Question

Find out the value of the following integral.
$\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{\sqrt[n]{\sec x}}{\sqrt[n]{\sec x}+\sqrt[n]{\operatorname{cosec}x}} \right)dx=}$
$\left( \text{a} \right)\text{ }\dfrac{\pi }{2}$
$\left( \text{b} \right)\text{ }\dfrac{\pi }{3}$
$\left( \text{c} \right)\text{ }\dfrac{\pi }{4}$
$\left( \text{d} \right)\text{ }\dfrac{\pi }{6}$

Answer 1

Hint : To solve the given question, we will use the king property of integration which says that,
$\int\limits_{a}^{b}{f\left( x \right)dx}=\int\limits_{a}^{b}{f\left( a+b-x \right)dx}$
Now, with the help of this, we will get another integral. We will assume that the value of these integrals is I. Then, we will add both of them. After simplifying, we will get the value of the integral 2I. On dividing the equation by 2, we will get the required result.

Complete step by step solution :
Before we find out the value of the given definite integral, we will first assume that the value of $\dfrac{\sqrt[n]{\sec x}}{\sqrt[n]{\sec x}+\sqrt[n]{\operatorname{cosec}x}}$ is f(x). Also, let the final value of the integral be I. Thus, we will get,
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{\sqrt[n]{\sec x}}{\sqrt[n]{\sec x}+\sqrt[n]{\operatorname{cosec}x}} \right)dx}.....\left( i \right)$
$f\left( x \right)=\dfrac{\sqrt[n]{\sec x}}{\sqrt[n]{\sec x}+\sqrt[n]{\operatorname{cosec}x}}......\left( ii \right)$
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{f\left( x \right)dx}.....\left( iii \right)$
Now, we know that we can write $\sqrt[r]{t}$ as ${{\left( t \right)}^{\dfrac{1}{r}}}.$ Thus, we will get,
$\Rightarrow f\left( x \right)=\dfrac{{{\left( \sec x \right)}^{\dfrac{1}{n}}}}{{{\left( \sec x \right)}^{\dfrac{1}{n}}}+{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{n}}}}$
Now, we will divide the numerator and denominator of the right-hand side with ${{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{n}}}.$ Thus, we will get,
$\Rightarrow f\left( x \right)=\dfrac{\dfrac{{{\left( \sec x \right)}^{\dfrac{1}{n}}}}{{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{n}}}}}{\dfrac{{{\left( \sec x \right)}^{\dfrac{1}{n}}}}{{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{n}}}}+\dfrac{{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{n}}}}{{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{n}}}}}$
$\Rightarrow f\left( x \right)=\dfrac{\dfrac{{{\left( \sec x \right)}^{\dfrac{1}{n}}}}{{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{n}}}}}{\dfrac{{{\left( \sec x \right)}^{\dfrac{1}{n}}}}{{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{n}}}}+1}......\left( iv \right)$
Now, we know that $\sec \theta =\dfrac{1}{\cos \theta }$ and $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }.$ Thus, we can say that,
$\dfrac{\sec x}{\operatorname{cosec}x}=\dfrac{\left( \dfrac{1}{\cos x} \right)}{\left( \dfrac{1}{\sin x} \right)}$
$\Rightarrow \dfrac{\sec x}{\operatorname{cosec}x}=\dfrac{\sin x}{\cos x}$
Now, we know that, $\dfrac{\sin \theta }{\cos \theta }=\tan \theta .$ Thus, we will get,
$\Rightarrow \dfrac{\sec x}{\operatorname{cosec}x}=\tan x......\left( v \right)$
From (iv) and (v), we have,
$f\left( x \right)=\dfrac{{{\left( \tan x \right)}^{\dfrac{1}{n}}}}{{{\left( \tan x \right)}^{\dfrac{1}{n}}}+1}.....\left( vi \right)$
Now, we will put the value of f(x) from (vi) to (iii). Thus, we will get,
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\left( \tan x \right)}^{\dfrac{1}{n}}}}{{{\left( \tan x \right)}^{\dfrac{1}{n}}}+1} \right)dx......\left( vii \right)}$
Now, to solve this integral, we will use the king property of integration which says that,
$\int\limits_{a}^{b}{g\left( x \right)dx}=\int\limits_{a}^{b}{g\left( a+b-x \right)dx}$
In our case, a = 0, $b=\dfrac{\pi }{2}$ and g(x) = f(x). Thus, we will get,
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\left( \tan \left( \dfrac{\pi }{2}-x \right) \right)}^{\dfrac{1}{n}}}}{{{\left( \tan \left( \dfrac{\pi }{2}-x \right) \right)}^{\dfrac{1}{n}}}+1} \right)dx}$
Now, $\tan \left( \dfrac{\pi }{2}-\theta \right)=\cot \theta .$ Thus, we will get,
$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\left( \cot x \right)}^{\dfrac{1}{n}}}}{{{\left( \cot x \right)}^{\dfrac{1}{n}}}+1} \right)dx}......\left( viii \right)$
Now, we will add the equations (vii) and (viii). On doing this, we will get,
$\Rightarrow I+I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\left( \tan x \right)}^{\dfrac{1}{n}}}}{{{\left( \tan x \right)}^{\dfrac{1}{n}}}+1} \right)dx}+\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\left( \cot x \right)}^{\dfrac{1}{n}}}}{{{\left( \cot x \right)}^{\dfrac{1}{n}}}+1} \right)dx}$
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\left( \tan x \right)}^{\dfrac{1}{n}}}}{{{\left( \tan x \right)}^{\dfrac{1}{n}}}+1}+\dfrac{{{\left( \cot x \right)}^{\dfrac{1}{n}}}}{{{\left( \cot x \right)}^{\dfrac{1}{n}}}+1} \right)dx}$
On taking the LCM, we will get the following equation.
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\left( \tan x \right)}^{\dfrac{1}{n}}}\left[ {{\left( \cot x \right)}^{\dfrac{1}{n}}}+1 \right]+{{\left( \cot x \right)}^{\dfrac{1}{n}}}\left[ {{\left( \tan x \right)}^{\dfrac{1}{n}}}+1 \right]}{\left[ {{\left( \tan x \right)}^{\dfrac{1}{n}}}+1 \right]\left[ {{\left( \cot x \right)}^{\dfrac{1}{n}}}+1 \right]} \right)dx}$
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\left( \tan x.\cot x \right)}^{\dfrac{1}{n}}}+{{\left( \tan x \right)}^{\dfrac{1}{n}}}+{{\left( \tan x.\cot x \right)}^{\dfrac{1}{n}}}+{{\left( \cot x \right)}^{\dfrac{1}{n}}}}{{{\left( \tan x.\cot x \right)}^{\dfrac{1}{n}}}+{{\left( \tan x \right)}^{\dfrac{1}{n}}}+{{\left( \cot x \right)}^{\dfrac{1}{n}}}+1} \right)dx}$
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\left( 1 \right)}^{\dfrac{1}{n}}}+{{\left( \tan x \right)}^{\dfrac{1}{n}}}+{{\left( 1 \right)}^{\dfrac{1}{n}}}+{{\left( \cot x \right)}^{\dfrac{1}{n}}}}{{{\left( 1 \right)}^{\dfrac{1}{n}}}+{{\left( \tan x \right)}^{\dfrac{1}{n}}}+{{\left( \cot x \right)}^{\dfrac{1}{n}}}+1} \right)dx}$
$\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{2+{{\left( \tan x \right)}^{\dfrac{1}{n}}}+{{\left( \cot x \right)}^{\dfrac{1}{n}}}}{2+{{\left( \tan x \right)}^{\dfrac{1}{n}}}+{{\left( \cot x \right)}^{\dfrac{1}{n}}}} \right)dx}$
Now, the numerator and denominator are the same on the right-hand side. So, we will cancel them. Thus, we will get the following equation. $\Rightarrow 2I=\int\limits_{0}^{\dfrac{\pi }{2}}{dx}$
Now, the integration of $\int{adx}=ax+c.$ Thus, we will get,
$\Rightarrow 2I=\left[ x+C \right]_{0}^{\dfrac{\pi }{2}}$
$\Rightarrow 2I=\left[ \left( \dfrac{\pi }{2}+C \right)-\left( 0+C \right) \right]$
$\Rightarrow 2I=\dfrac{\pi }{2}+C-0-C$
$\Rightarrow 2I=\dfrac{\pi }{2}$
On dividing by 2, we will get,
$\Rightarrow I=\dfrac{\pi }{4}.......\left( ix \right)$
From (i) and (ix), we have,
$\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{\sqrt[n]{\sec x}}{\sqrt[n]{\sec x}+\sqrt[n]{\operatorname{cosec}x}} \right)dx=}\dfrac{\pi }{4}$
Hence, the option (c) is the right option.

Note : We can directly apply the king property without converting f(x) in terms of tan x. Thus, we have,
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\left( \sec x \right)}^{\dfrac{1}{n}}}}{{{\left( \sec x \right)}^{\dfrac{1}{n}}}+{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{n}}}} \right)dx......\left( i \right)}$
$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\left( \sec \left( \dfrac{\pi }{2}-x \right) \right)}^{\dfrac{1}{n}}}}{{{\left( \sec \left( \dfrac{\pi }{2}-x \right) \right)}^{\dfrac{1}{n}}}+{{\left( \operatorname{cosec}\left( \dfrac{\pi }{2}-x \right) \right)}^{\dfrac{1}{n}}}} \right)dx}$
$\Rightarrow I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \dfrac{{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{n}}}}{{{\left( \operatorname{cosec}x \right)}^{\dfrac{1}{n}}}+{{\left( \sec x \right)}^{\dfrac{1}{n}}}} \right)dx}.......\left( ii \right)$
On adding (i) and (ii), we will get,
$2I=\int{dx}$
$\Rightarrow 2I=\left[ x \right]_{0}^{\dfrac{\pi }{2}}$
$\Rightarrow 2I=\dfrac{\pi }{2}$
$\Rightarrow I=\dfrac{\pi }{4}$