Question

If $\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cot x}{\cot x+\csc x}dx}=m\left( \pi +n \right)$, then ‘mn’ is equal to?
(a) -1
(b) 1
(c) $\dfrac{1}{2}$
(d) $-\dfrac{1}{2}$

Answer 1

Hint: Don’t use any property related with definite integral. Convert the given relation in ‘cot x’ and ‘csc x’ to ‘sin x’ and ‘cos x’. Now, try to simplify it further to get the value of the integral and hence by comparison, find ‘mn’.

Complete step by step answer:
Let $I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cot x}{\cot x+\csc x}dx}$ ………………… (i)
We can convert the above integral in terms of ‘sin’ and ‘cos’ by using the relation,
\[\cot \theta =\dfrac{\cos \theta }{\sin \theta }\text{,csc}\theta \text{=}\dfrac{1}{\sin \theta }\]
Hence, equation(i) can be written as
\[\begin{align}
  & I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{\cos x}{\sin x}+\dfrac{1}{\sin x}}}dx \\
 & I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\dfrac{\cos x}{\sin x}}{\dfrac{\cos x+1}{\sin x}}}dx \\
\end{align}\]
Cancelling the like terms, we get
\[I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cos x}{1+\cos x}}dx\]……… (ii)
Now, we can use identity of \[\cos 2\theta \] in terms of \[\tan \theta \] which is given as,
\[\cos 2\theta =\dfrac{1-{{\tan }^{2}}\theta }{1+{{\tan }^{2}}\theta }\] ……………….. (iii)
Hence, equation (ii) can be written as
\[I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\dfrac{1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)}{1+{{\tan }^{2}}\left( \dfrac{x}{2} \right)}}{1+\dfrac{1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)}{1+{{\tan }^{2}}\left( \dfrac{x}{2} \right)}}dx}\]
Now, simplifying the above relation, we get
\[I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)}{1+{{\tan }^{2}}\left( \dfrac{x}{2} \right)}\times \dfrac{1+{{\tan }^{2}}\left( \dfrac{x}{2} \right)}{1+{{\tan }^{2}}\dfrac{x}{2}+1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)}}dx\]
Or
\[I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)}{2}\times \dfrac{1+{{\tan }^{2}}\left( \dfrac{x}{2} \right)}{1+{{\tan }^{2}}\left( \dfrac{x}{2} \right)}}dx\]
Now, replace \[1+{{\tan }^{2}}\left( \dfrac{x}{2} \right)\] by \[{{\sec }^{2}}\left( \dfrac{x}{2} \right)\]by the trigonometric identity \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \], we get
\[I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{1-{{\tan }^{2}}\left( \dfrac{x}{2} \right)}{2}\times \dfrac{{{\sec }^{2}}\left( \dfrac{x}{2} \right)}{1+{{\tan }^{2}}\left( \dfrac{x}{2} \right)}dx}\]………………………. (iv)
Now, we can suppose $\tan \dfrac{x}{2}$ as ‘t’ and differentiate it to get relation in ‘dt’ and ‘dx’ in following way:
$\tan \dfrac{x}{2}=t,$
$\dfrac{1}{2}{{\sec }^{2}}\dfrac{x}{2}dx=dt$
${{\sec }^{2}}\dfrac{x}{2}dx=2dt$
Now, we can change limits of the integral given in equation (iv), by using relation $t=\tan \left( \dfrac{x}{2} \right)$
Upper limit w.r.t$x=\dfrac{\pi }{2}$
Upper limit w.r.t $t=\tan \dfrac{\pi }{4}=1$
Lower limit w.r.t x=0
Lower limit w.r.t $t=\tan \left( \dfrac{0}{2} \right)=0$
Hence, integral ‘I’ w.r.t dt can be written as
$I=\int\limits_{0}^{1}{\dfrac{1-{{t}^{2}}}{2\left( 1+{{t}^{2}} \right)}\times 2dt}$
$I=\int\limits_{0}^{1}{\dfrac{1-{{t}^{2}}}{1+{{t}^{2}}}dt}$
Now adding and subtracting ‘1’ in the numerator of above equation, we get
$I=\int\limits_{0}^{1}{\dfrac{1-{{t}^{2}}-1+1}{1+{{t}^{2}}}dt}$
$I=\int\limits_{0}^{1}{\dfrac{2-\left( 1+{{t}^{2}} \right)}{1+{{t}^{2}}}dt}$
$I=\int\limits_{0}^{1}{\dfrac{2}{1+{{t}^{2}}}dt}-\int\limits_{0}^{1}{1dt}$
$I=2\int\limits_{0}^{1}{\dfrac{1}{1+{{t}^{2}}}dt}-\int\limits_{0}^{1}{1dt}$
Now, we can use relation, $\int{\dfrac{1}{{{x}^{2}}+{{a}^{2}}}}dx=\dfrac{1}{a}{{\tan }^{-1}}\left( \dfrac{x}{a} \right)$ and$\int{1dx=x}$.
Hence, I can be given as
\[I=2\left. {{\tan }^{-1}}t \right|_{0}^{1}-\left. t \right|_{0}^{1}\]
Applying the limits, we get
$I=2\left( {{\tan }^{-1}}1-{{\tan }^{-1}}0 \right)-\left( 1-0 \right)$
Now, we know values of ${{\tan }^{-1}}1$and ${{\tan }^{-1}}0$ which are given as $\dfrac{\pi }{4}$and 0. Hence, we get
$I=2\left( \dfrac{\pi }{4}-0 \right)-1$
$I=\dfrac{\pi }{2}-1=\dfrac{1}{2}\left( \pi -2 \right)$
The integral in the problem is given as $m\left( \pi +n \right)$. Hence, by comparing the calculated and given integral, we get values of m and n as
$m=\dfrac{1}{2}$ and $n=-2$
Hence,
$mn=-2\times \dfrac{1}{2}=-1$.
Therefore, option (a) is the correct answer.

Note: Another approach for solving the integral would be that we can multiply the given integral by $\csc x-\cot x$ in numerator and denominator both. Hence, integral becomes
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\dfrac{\cot x}{\csc x+\cot x}\times \left( \dfrac{\csc x-\cot x}{\csc x-\cot x} \right)}$
$I=\int\limits_{0}^{\dfrac{\pi }{2}}{\left( \cot x\csc x-{{\cot }^{2}}x \right)dx}$
Where ${{\csc }^{2}}x-{{\cot }^{2}}x=1$
Now, integral it further by changing the expression in terms of ‘sin x’ and ‘cos x’.