Question

The value of the integral $\int{\dfrac{dx}{\cos x-\sin x}}$ is equal to:
(a) $-\dfrac{1}{\sqrt{2}}\log \left| \tan \left[ \dfrac{x}{2}-\dfrac{\pi }{8} \right] \right|+C$
(b) $\dfrac{1}{\sqrt{2}}\log \left| \cot \left[ \dfrac{x}{2} \right] \right|+C$
(c) $\dfrac{1}{\sqrt{2}}\log \left| \tan \left[ \dfrac{x}{2}-\dfrac{3\pi }{8} \right] \right|+C$
(d) $\dfrac{1}{\sqrt{2}}\log \left| \tan \left[ \dfrac{x}{2}+\dfrac{\pi }{8} \right] \right|+C$

Answer 1

Hint: Multiplying and dividing the given integral with $\sqrt{2}$ we get $\int{\dfrac{\sqrt{2}dx}{\sqrt{2}\left( \cos x-\sin x \right)}}$. Now, we can rewrite the above equation as $\int{\dfrac{dx}{\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\cos x-\dfrac{1}{\sqrt{2}}\sin x \right)}}$. The denominator in this equation apart from $\sqrt{2}$ can be written as $\cos \left( \dfrac{\pi }{4}+x \right)$so the integral is reduced to $\int{\dfrac{dx}{\sqrt{2}\left( \cos \left( \dfrac{\pi }{4}+x \right) \right)}}$. Solving this integral we get $\dfrac{1}{\sqrt{2}}\ln \left( \sec \left( \dfrac{\pi }{4}+x \right)+\tan \left( \dfrac{\pi }{4}+x \right) \right)$. Using trigonometric properties simplify this new expression.

Complete step-by-step answer:
The integral given in the question is:
$\int{\dfrac{dx}{\cos x-\sin x}}$
Multiplying and dividing $\sqrt{2}$ in the above integral we get,
$\int{\dfrac{\sqrt{2}dx}{\sqrt{2}\left( \cos x-\sin x \right)}}$
Rewriting the above integral we get,
$\int{\dfrac{dx}{\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\cos x-\dfrac{1}{\sqrt{2}}\sin x \right)}}$………..Eq. (1)
The denominator in the above integral apart from $\sqrt{2}$ is the expansion of $\cos \left( \dfrac{\pi }{4}+x \right)$.
$\cos \left( \dfrac{\pi }{4}+x \right)=\cos \dfrac{\pi }{4}\cos x-\sin \dfrac{\pi }{4}\sin x$
And we know that the trigonometric value of $\sin \dfrac{\pi }{4}\And \cos \dfrac{\pi }{4}$ is $\dfrac{1}{\sqrt{2}}$ so substituting this relation in the above equation we get,
$\cos \left( \dfrac{\pi }{4}+x \right)=\dfrac{1}{\sqrt{2}}\cos x-\dfrac{1}{\sqrt{2}}\sin x$
Using these relations in the eq. (1) we get,
$\begin{aligned} & \int{\dfrac{dx}{\sqrt{2}\left( \cos \left( \dfrac{\pi }{4}+x \right) \right)}} \\\ & =\int{\dfrac{1}{\sqrt{2}}\left( \sec \left( x+\dfrac{\pi }{4} \right) \right)}dx \\\ \end{aligned}$
We know that the integration of $\sec x$ with respect to x is $\ln \left( \sec x+\tan x \right)+C$.
$\dfrac{1}{\sqrt{2}}\ln \left| \left( \sec \left( x+\dfrac{\pi }{4} \right)+\tan \left( x+\dfrac{\pi }{4} \right) \right) \right|+C$ ………. Eq. (2)
$=\dfrac{1}{\sqrt{2}}\ln \left| \left( \dfrac{1}{\cos \left( x+\dfrac{\pi }{4} \right)}+\dfrac{\sin \left( x+\dfrac{\pi }{4} \right)}{\cos \left( x+\dfrac{\pi }{4} \right)} \right) \right|+C$
Taking L.C.M of $\cos \left( \dfrac{\pi }{4}+x \right)$ in the logarithmic expression we get,
$=\dfrac{1}{\sqrt{2}}\ln \left| \left( \dfrac{1+\sin \left( x+\dfrac{\pi }{4} \right)}{\cos \left( x+\dfrac{\pi }{4} \right)} \right) \right|+C$
In the above expression we can write $\sin \left( x+\dfrac{\pi }{4} \right)\And \cos \left( x+\dfrac{\pi }{4} \right)$ in terms of $\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)$ as follows:
$\begin{aligned} & \dfrac{1}{\sqrt{2}}\ln \left| \left( \dfrac{1+\dfrac{2\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)}{1+{{\tan }^{2}}\left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)}}{\dfrac{1-{{\tan }^{2}}\left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)}{1+{{\tan }^{2}}\left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)}} \right) \right|+C \\\ & =\dfrac{1}{\sqrt{2}}\ln \left| \left( \dfrac{1+{{\tan }^{2}}\left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)+2\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)}{1-{{\tan }^{2}}\left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)} \right) \right|+C \\\ \end{aligned}$
In the above expression, the numerator in the logarithm is the expansion of ${{\left( 1+\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right) \right)}^{2}}$.

& \dfrac{1}{\sqrt{2}}\ln \left| \left( \dfrac{{{\left( 1+\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right) \right)}^{2}}}{1-{{\tan }^{2}}\left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)} \right) \right|+C \\\ & =\dfrac{1}{\sqrt{2}}\ln \left| \left( \dfrac{{{\left( 1+\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right) \right)}^{2}}}{\left( 1-\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)\left( 1+\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right) \right) \right)} \right) \right|+C \\\ \end{aligned}$$ In the above expression $\left( 1+\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right) \right)$ will be cancelled out in the numerator and denominator of the expression in logarithm. $\dfrac{1}{\sqrt{2}}\ln \left| \left( \dfrac{1+\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)}{1-\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)} \right) \right|+C$ Rewriting the above expression as: $\begin{aligned} & \dfrac{1}{\sqrt{2}}\ln {{\left| \left( \dfrac{1-\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)}{1+\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)} \right) \right|}^{-1}}+C \\\ & =-\dfrac{1}{\sqrt{2}}\ln \left| \left( \dfrac{1-\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)}{1+\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)} \right) \right|+C \\\ \end{aligned}$ We can write 1 in the above expression as $\tan \dfrac{\pi }{4}$. $-\dfrac{1}{\sqrt{2}}\ln \left| \left( \dfrac{\tan \dfrac{\pi }{4}-\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)}{1+\tan \dfrac{\pi }{4}\tan \left( \dfrac{x}{2}+\dfrac{\pi }{8} \right)} \right) \right|+C$ The expression in the logarithm is the expansion of $\tan \left( \dfrac{\pi }{4}-\dfrac{x}{2}-\dfrac{\pi }{8} \right)$. $\begin{aligned} & -\dfrac{1}{\sqrt{2}}\ln \left| \left( \tan \left( \dfrac{\pi }{4}-\dfrac{x}{2}-\dfrac{\pi }{8} \right) \right) \right|+C \\\ & =-\dfrac{1}{\sqrt{2}}\ln \left| \left( \tan \left( \dfrac{\pi }{8}-\dfrac{x}{2} \right) \right) \right|+C \\\ \end{aligned}$ We can write the above expression as: $-\dfrac{1}{\sqrt{2}}\ln \left| \left( -\tan \left( \dfrac{x}{2}-\dfrac{\pi }{8} \right) \right) \right|+C$ $-\dfrac{1}{\sqrt{2}}\ln \left| \left( \tan \left( \dfrac{x}{2}-\dfrac{\pi }{8} \right) \right) \right|+C$ In the above expression “ln” is the abbreviation for log base “e” so we can write “ln” as log. $-\dfrac{1}{\sqrt{2}}\log \left| \tan \left( \dfrac{x}{2}-\dfrac{\pi }{8} \right) \right|+C$ From the above solution, the value of the given integral is $-\dfrac{1}{\sqrt{2}}\log \left| \tan \left( \dfrac{x}{2}-\dfrac{\pi }{8} \right) \right|+C$. Hence, the correct option is (a). Note: You might think we can stop at eq. (2) why we have gone further. The answer is the options in the question are not in the form of eq. (2) so we have simplified it further to reach the expression that the options contain.