Question

The value of $\dfrac{{{{\pi}^2}}}{{\ln 3}}\int \nolimits_{7/6}^{5/6} sec\left( {\pi x} \right)dx$ is

Answer 1

Hint : First consider $\pi x$ as t. and then differentiate t with respect to x. Substitute the obtained values accordingly in the given trigonometric expression. As we have changed the given expression, the limits also will be changed according to the values of x and t. And then find the integration of secant function using the below mentioned formula.
Formula used:
Integration of $\int secdx = \log \left| {secx + \tan x} \right|$

Complete step-by-step answer :
We are given to find the value of $\dfrac{{{{\pi}^2}}}{{\ln 3}}\int \nolimits_{7/6}^{5/6} sec\left( {\pi x} \right)dx$
Let us first consider $\pi x$ as t.
$t = \pi x$
On differentiating the above equation with respect to x, we get
$\dfrac{d}{{dx}}t = \dfrac{d}{{dx}}\left( {\pi x} \right)$
$\Rightarrow \dfrac{{dt}}{{dx}} = {\pi}\dfrac{{dx}}{{dx}} = {\pi}$
$\Rightarrow dx = \dfrac{{dt}}{{\pi}}$
Substituting the obtained value of $dx$ and $\pi x$ , we get
$\Rightarrow \dfrac{{{{\pi}^2}}}{{\ln 3}}\int \nolimits_{7{\pi}/6}^{5{\pi}/6} \dfrac{{sect}}{{\pi}}dt$ .
The limits change when x is replaced by the terms of t as $x = \dfrac{5}{6} \Rightarrow t = \dfrac{{5{\pi}}}{6}$ and $x = \dfrac{7}{6} \Rightarrow t = \dfrac{{7{\pi}}}{6}$ .
Taking out ${\pi}$ which is in the denominator, we get
$\Rightarrow \dfrac{{{{\pi}^2}}}{{\ln 3}} \times \dfrac{1}{{\pi}}\int \nolimits_{7{\pi}/6}^{5{\pi}/6} sectdt = \dfrac{{\pi}}{{\ln 3}}\int \nolimits_{7{\pi}/6}^{5{\pi}/6} sectdt$
Now apply the integration formula $\int secdx = \log \left| {secx + \tan x} \right|$ as x is equal to t.
$\Rightarrow \dfrac{{\pi}}{{\ln 3}}\left. {\log \left| {secx + \tan x} \right|} \right|_{7{\pi}/6}^{5{\pi}/6}$
Applying the limits, we get
$\Rightarrow \dfrac{{\pi}}{{\ln 3}}\log \left| {\left( {sec\;\dfrac{{5{\pi}}}{6} + \tan \;\dfrac{{5{\pi}}}{6}} \right) - \left( {sec\;\dfrac{{7{\pi}}}{6} + \tan\; \dfrac{{7{\pi}}}{6}} \right)} \right|$
$\dfrac{{5{\pi}}}{6}$ is also equal to 150 degrees and $\dfrac{{7{\pi}}}{6}$ is also equal to 210 degrees.
$\Rightarrow \dfrac{{\pi}}{{\ln 3}}\log \left| {\left( {sec\;{{150}^ \circ } + \tan \;{{150}^ \circ }} \right) - \left( {sec\;{{210}^ \circ } + \tan \;{{210}^ \circ }} \right)} \right|$
The value of $sec\;{150^ \circ } = - \dfrac{2}{{\sqrt 3 }}$ , $\tan\;{150^ \circ } = - \dfrac{1}{{\sqrt 3 }}$ , $sec\;{210^ \circ } = - \dfrac{2}{{\sqrt 3 }}$ and $\tan\;{210^ \circ } = \dfrac{1}{{\sqrt 3 }}$
Therefore,
$\Rightarrow \dfrac{{\pi}}{{\ln 3}}\log \left| {\left( {sec\;{{150}^ \circ } + \tan \;{{150}^ \circ }} \right) - \left( {sec\;{{210}^ \circ } + \tan\;{{210}^ \circ }} \right)} \right| = \dfrac{{\pi}}{{\ln 3}}\log \left| {\left( {\dfrac{{ - 2}}{{\sqrt 3 }} - \dfrac{1}{{\sqrt 3 }}} \right) - \left( {\dfrac{{ - 2}}{{\sqrt 3 }} + \dfrac{1}{{\sqrt 3 }}} \right)} \right|$
$\Rightarrow \dfrac{{\pi}}{{\ln 3}}\log \left| {\dfrac{{ - 2}}{{\sqrt 3 }} - \dfrac{1}{{\sqrt 3 }} - \dfrac{{ - 2}}{{\sqrt 3 }} - \dfrac{1}{{\sqrt 3 }}} \right| = \dfrac{{\pi}}{{\ln 3}}\log \left| { - \dfrac{2}{{\sqrt 3 }}} \right| = \dfrac{{\pi}}{{\ln 3}}\log \dfrac{2}{{\sqrt 3 }}$
Therefore, the value of $\dfrac{{{{\pi}^2}}}{{\ln 3}}\int \nolimits_{7/6}^{5/6} sec\left( {\pi x} \right)dx$ is $\dfrac{{\pi}}{{\ln 3}}\log \dfrac{2}{{\sqrt 3 }}$
So, the correct answer is “$\dfrac{{\pi}}{{\ln 3}}\log \dfrac{2}{{\sqrt 3 }}$ ”.

Note : We can also write secant as the reciprocal of cosine and then solve the integration. Here we have taken the integration of secant in terms of “log” which is defined for base 10. We can also consider it as “ln” which is defined for base ‘e’, e is the exponential function. Be careful while calculating the differentiation and integration.