Question

How do you find the definite integral of $\tan xdx$ from $\left[ \dfrac{\pi }{4},\pi \right]$?

Answer 1

This type of question is based on the concept of definite integral. First, we have to simplify the given function as $\tan x=\dfrac{\sin x}{\cos x}$. Then, consider u=cosx and find du. Substitute these to find the definite integral. Do necessary calculations. And then apply the limits to find the final solution.

Complete step by step answer:
According to the question, we are asked to find the definite integral of $\tan xdx$ from $\left[ \dfrac{\pi }{4},\pi \right]$.
We have been given the equation is $\tan xdx$ --------(1)
We know that $\tan x=\dfrac{\sin x}{\cos x}$.
Therefore, $\tan xdx=\dfrac{\sin x}{\cos x}dx$
Now, let us find the definite integral.
$\int_{\dfrac{\pi }{4}}^{\pi }{\tan xdx=}\int_{\dfrac{\pi }{4}}^{\pi }{\dfrac{\sin x}{\cos x}dx}$ -------(2)
Let us now substitute u=cosx
But here $x\in \left[ \dfrac{\pi }{4},\pi \right]$.
When $x=\dfrac{\pi }{4}$, $u=\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}$
Similarly, when $x=\pi$,$u=\sin \pi =0$
Therefore, the limits with respect to x is $\left[ \dfrac{1}{\sqrt{2}},0 \right]$
Let us now differentiate ‘u’ with respect to ‘x’.
$\dfrac{du}{dx}=\dfrac{d}{dx}\left( \cos x \right)$
We know that $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$.
$\Rightarrow \dfrac{du}{dx}=-\sin x$
$\therefore du=-\sin xdx$ --------(3)
Substituting u and (3) in equation (2), we get
$\int_{\dfrac{\pi }{4}}^{\pi }{\tan xdx=}\int_{\dfrac{1}{\sqrt{2}}}^{0}{\dfrac{-1}{u}du}$
$\Rightarrow \int_{\pi }^{\dfrac{\pi }{4}}{\tan xdx=}-\int_{\dfrac{1}{\sqrt{2}}}^{0}{\dfrac{1}{u}du}$
We know that $\int{\dfrac{1}{x}dx=\log x}$, we get
$\Rightarrow \int_{\dfrac{\pi }{4}}^{\pi }{\tan xdx=}-\left[ \log u \right]_{\dfrac{1}{\sqrt{2}}}^{0}$
$\Rightarrow \int_{\dfrac{\pi }{4}}^{\pi }{\tan xdx=}-\left[ \log 0-\log \left( \dfrac{1}{\sqrt{2}} \right) \right]$
We know that log0=0.
$\Rightarrow \int_{\dfrac{\pi }{4}}^{\pi }{\tan xdx=}-\left[ -\log \left( \dfrac{1}{\sqrt{2}} \right) \right]$
$\therefore \int_{\dfrac{\pi }{4}}^{\pi }{\tan xdx=}\log \left( \dfrac{1}{\sqrt{2}} \right)$

Hence, find the definite integral of $\tan xdx$ from $\left[ \dfrac{\pi }{4},\pi \right]$ is $\log \left( \dfrac{1}{\sqrt{2}} \right)$

Note:
We can further simplify the answer using logarithmic properties such as
$\log \left( \dfrac{1}{x} \right)=-\log x$ and $\log \left( {{x}^{n}} \right)=n\log x$.
Therefore,
$\log \left( \dfrac{1}{\sqrt{2}} \right)=-\log \left( \sqrt{2} \right)$
$\Rightarrow \log \left( \dfrac{1}{\sqrt{2}} \right)=-\log \left( {{2}^{\dfrac{1}{2}}} \right)$
$\therefore \log \left( \dfrac{1}{\sqrt{2}} \right)=-\dfrac{1}{2}\log \left( 2 \right)$
Hence the final solution of the definite integral of $\tan xdx$ from $\left[ \dfrac{\pi }{4},\pi \right]$ is $-\dfrac{1}{2}\log \left( 2 \right)$.
For this type of questions, we should use trigonometric identities and simplify the given function of finding the definite integral. We should avoid calculation mistakes based on sign conventions. We should also know the differentiation of trigonometric functions for easy calculations. We should not forget to convert the limits of the given function. If we forget to change the value of the limits, the answer will be incorrect. We can also solve this question by not substituting u=cosx. But instead, we can directly integrate tanxdx.