Question

Evaluate the following definite integral:
$\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cot xdx}$

Answer 1

Hint: In this question we have to solve the given definite integral. We will first convert the function given as $\cot x=\dfrac{\cos x}{\sin x}$ . Then by using the integration formula we will solve the integral. Then after solving we will substitute the limits given and then simplifying the obtained equation we will get the desired answer.

Complete step by step solution:
We have been given a definite integral $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cot xdx}$.
We have to solve the given integral.
$= \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cot xdx}$
Now, we know that $\cot x=\dfrac{\cos x}{\sin x}$.
Now, substituting the values we will get
$= \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\cos x}{\sin x}dx}$
Now, we know that $\int{\dfrac{f'(x)}{f(x)}dx=\log \left| f\left( x \right) \right|}$
Here we have $f\left( x \right)=\sin xdx\text{ and }f'\left( x \right)=\cos x$
So by applying the formula to the integral we will get
$= \log \left| \sin x \right|_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}$
Now, putting the limits we will get
$= \log \left| \sin \dfrac{\pi }{2} \right|-\log \left| \sin \dfrac{\pi }{4} \right|$
Now, we know that $\sin \dfrac{\pi }{2}=1\text{ and }\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$
Now, substituting the values we will get
$\begin{align}
  & = \log \left| 1 \right|-\log \left| \dfrac{1}{\sqrt{2}} \right| \\
 & = \dfrac{1}{2}\log 2 \\
\end{align}$
Hence above is the required value of definite integral.

Note: Alternatively we can also solve the integral $\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\cos x}{\sin x}dx}$ by using the substitution method. We will substitute the function $\sin x=u$ then we get $du=\cos xdx$. Then we will substitute the values in the given integral, then we will get
$= \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{1}{u}du}$
Now, we know that $\int{\dfrac{1}{f(x)}dx=\log \left| f\left( x \right) \right|}$
Then simplifying the above obtained equation we will get
$= \left| \log u \right|_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}$
Now, again substituting $\sin x=u$we will get
$= \left| \log \sin x \right|_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}$
Now, putting the limits we will get
$= \log \left| \sin \dfrac{\pi }{2} \right|-\log \left| \sin \dfrac{\pi }{4} \right|$
Now, we know that $\sin \dfrac{\pi }{2}=1\text{ and }\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$
Now, substituting the values we will get
$\begin{align}
  & = \log \left| 1 \right|-\log \left| \dfrac{1}{\sqrt{2}} \right| \\
 & = \dfrac{1}{2}\log 2 \\
\end{align}$
Hence above is the required value of definite integral.