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Question: How do you find the integral of \[\int{\csc x dx}\] from \(\dfrac{\pi}{2}\) to \(\pi\)?...

How do you find the integral of ∫csc⁡xdx\int{\csc x dx} from π2\dfrac{\pi}{2} to π\pi?

Explanation

Solution

To find the integral in the given interval of the given trigonometric function is done by integrating the function first and then putting the intervals on the integrated trigonometric functions and subtracting the values of function in the given interval (that is from pi/2 to pi) in the prescribed order. But as we know that cosecant function cannot be integrated directly so we will first convert it into a form which can be integrated easily or is part of a formula and can then proceed with applying the intervals.

Complete step-by-step solution:
Cosecant is one of the trigonometric functions. Cosecant and sine functions are inverse of each other. Sine function shows sinusoidal graph whereas cosecant has neither maximum nor minimum.
Integration refers to a process in which an infinitesimal data is used to describe a larger concept like area, volume, etc. And also integration and differentiation are reversed each other. We can understand through the following example.
For example- derivative of sin⁡x\sin x is ddx(sin⁡x)=cos⁡x\dfrac{d}{dx}(\sin x)=\cos x
And the integration of cos⁡x\cos x is ∫cos⁡x=sin⁡x+C\int{\cos x=\sin x+C} (where C is a constant)
Therefore, they are reverse processes of each other.
According to the question, we have to integrate cosecant function in the given interval, we can re-write the entire question as:
∫π2πcsc⁡xdx\int\limits_{\dfrac{\pi }{2}}^{\pi }{\csc xdx}
We will start solving the integral by multiplying and dividing by csc⁡x−cot⁡x\csc x-\cot x, we get
⇒∫π2πcsc⁡x×csc⁡x−cot⁡xcsc⁡x−cot⁡xdx\Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\pi }{\csc x\times \dfrac{\csc x-\cot x}{\csc x-\cot x}dx}
⇒∫π2πcsc⁡2x−csc⁡xcot⁡xcsc⁡x−cot⁡xdx\Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\pi }{\dfrac{{{\csc }^{2}}x-\csc x\cot x}{\csc x-\cot x}dx}
Rearranging we get,
⇒∫π2π−csc⁡xcot⁡x+csc⁡2xcsc⁡x−cot⁡xdx\Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\pi }{\dfrac{-\csc x\cot x+{{\csc }^{2}}x}{\csc x-\cot x}dx}
⇒∫π2π−csc⁡xcot⁡xdx+csc⁡2xdxcsc⁡x−cot⁡x\Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\pi }{\dfrac{-\csc x\cot xdx+{{\csc }^{2}}xdx}{\csc x-\cot x}}
We know that, ddx(csc⁡x)=−csc⁡xcot⁡x\dfrac{d}{dx}(\csc x)=-\csc x\cot x and ddx(cot⁡x)=−csc⁡2x\dfrac{d}{dx}(\cot x)=-{{\csc }^{2}}x, we will be substituting these in the integral above. On substitution, we get,
⇒∫π2πd(csc⁡x)−d(cot⁡x)csc⁡x−cot⁡x\Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\pi }{\dfrac{d(\csc x)-d(\cot x)}{\csc x-\cot x}}
⇒∫π2πd(csc⁡x−cot⁡x)csc⁡x−cot⁡x\Rightarrow \int\limits_{\dfrac{\pi }{2}}^{\pi }{\dfrac{d(\csc x-\cot x)}{\csc x-\cot x}}
So, we can see, the numerator is nothing but the derivative of the denominator and we know the formula that,
∫f′(x)f(x)=log⁡∣f(x)∣+c\int{\dfrac{f'(x)}{f(x)}=\log |f(x)|+c}
Applying this formula in the integral we have and then applying the intervals as well, we get,
⇒[log⁡∣csc⁡x−cot⁡x∣]π2π\Rightarrow \left[ \log |\csc x-\cot x| \right]_{\dfrac{\pi }{2}}^{\pi }
Applying the intervals, we have,
⇒log⁡(csc⁡π−cot⁡π)−log⁡(csc⁡π2−cot⁡π2)\Rightarrow \log (\csc \pi -\cot \pi )-\log (\csc \dfrac{\pi }{2}-\cot \dfrac{\pi }{2})
But we know that csc⁡π\csc \pi and cot⁡π\cot \pi have ‘not defined’ values. So, in the given intervals the integral is not defined.
Therefore, the integral in the given interval is not defined.

Note: Always remember that integration and differentiation are the reverse of each other and this can prove useful in integration questions. While applying formulae, be sure of the conditions under which the formula is applicable only then proceed with the integration.