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Question: Evaluate the following definite integral: \(\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cot x...

Evaluate the following definite integral:
∫π4π2cot⁡xdx\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cot xdx}

Explanation

Solution

In this question we have to solve the given definite integral. We will first convert the function given as cot⁡x=cos⁡xsin⁡x\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 ∫π4π2cot⁡xdx\int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cot xdx}.
We have to solve the given integral.
=∫π4π2cot⁡xdx= \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\cot xdx}
Now, we know that cot⁡x=cos⁡xsin⁡x\cot x=\dfrac{\cos x}{\sin x}.
Now, substituting the values we will get
=∫π4π2cos⁡xsin⁡xdx= \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{\cos x}{\sin x}dx}
Now, we know that ∫f′(x)f(x)dx=log⁡∣f(x)∣\int{\dfrac{f'(x)}{f(x)}dx=\log \left| f\left( x \right) \right|}
Here we have f(x)=sin⁡xdx and f′(x)=cos⁡xf\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⁡∣sin⁡x∣π4π2= \log \left| \sin x \right|_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}
Now, putting the limits we will get
=log⁡∣sin⁡π2∣−log⁡∣sin⁡π4∣= \log \left| \sin \dfrac{\pi }{2} \right|-\log \left| \sin \dfrac{\pi }{4} \right|
Now, we know that sin⁡π2=1 and sin⁡π4=12\sin \dfrac{\pi }{2}=1\text{ and }\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}
Now, substituting the values we will get
=log⁡∣1∣−log⁡∣12∣ =12log⁡2 \begin{aligned} & = \log \left| 1 \right|-\log \left| \dfrac{1}{\sqrt{2}} \right| \\\ & = \dfrac{1}{2}\log 2 \\\ \end{aligned}
Hence above is the required value of definite integral.

Note: Alternatively we can also solve the integral ∫π4π2cos⁡xsin⁡xdx\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\sin x=u then we get du=cos⁡xdxdu=\cos xdx. Then we will substitute the values in the given integral, then we will get
=∫π4π21udu= \int\limits_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}{\dfrac{1}{u}du}
Now, we know that ∫1f(x)dx=log⁡∣f(x)∣\int{\dfrac{1}{f(x)}dx=\log \left| f\left( x \right) \right|}
Then simplifying the above obtained equation we will get
=∣log⁡u∣π4π2= \left| \log u \right|_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}
Now, again substituting sin⁡x=u\sin x=uwe will get
=∣log⁡sin⁡x∣π4π2= \left| \log \sin x \right|_{\dfrac{\pi }{4}}^{\dfrac{\pi }{2}}
Now, putting the limits we will get
=log⁡∣sin⁡π2∣−log⁡∣sin⁡π4∣= \log \left| \sin \dfrac{\pi }{2} \right|-\log \left| \sin \dfrac{\pi }{4} \right|
Now, we know that sin⁡π2=1 and sin⁡π4=12\sin \dfrac{\pi }{2}=1\text{ and }\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}
Now, substituting the values we will get
=log⁡∣1∣−log⁡∣12∣ =12log⁡2 \begin{aligned} & = \log \left| 1 \right|-\log \left| \dfrac{1}{\sqrt{2}} \right| \\\ & = \dfrac{1}{2}\log 2 \\\ \end{aligned}
Hence above is the required value of definite integral.