Question

Evaluate $\int (\cos x-\sin x)dx$

• A. $\left(\frac{1}{\sqrt{2}}\right)\log \left|\tan \left[\left(\frac{x}{2}\right)–\left(\frac{\pi }{8}\right)\right]\right|+c$
• B. $\left(\frac{1}{\sqrt{2}}\right)\log \left|\cot \left(\frac{x}{2}\right)\right|+c$
• C. $\left(\frac{1}{\sqrt{2}}\right)\log \left|\cot \left[\left(\frac{x}{2}\right)–\left(\frac{3\pi }{8}\right)\right]\right|+c$
• D. $\left(\frac{1}{\sqrt{2}}\right)\log \left|\tan \left[\left(\frac{x}{2}\right)+\left(\frac{3\pi }{8}\right)\right]\right|+c$

Answer 1

The correct option is D

$\left(\frac{1}{\sqrt{2}}\right)\log \left|\tan \left[\left(\frac{x}{2}\right)+\left(\frac{3\pi }{8}\right)\right]\right|+c$

Explanation for the correct option

Integrating the given integral

Given, $\int (\cos x-\sin x)dx$

Solving the integral,

$\begin{array}{l}\begin{array}{l}\int (\cos x-\sin x)dx=\frac{1}{\sqrt{2}}\int \frac{dx}{\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x}\\ =\frac{1}{\sqrt{2}}\int \frac{dx}{\cos {\displaystyle \frac{\mathrm{\pi }}{4}}.\cos x-\sin {\displaystyle \frac{\mathrm{\pi }}{4}}.\sin x}\\ =\frac{1}{\sqrt{2}}\int \frac{dx}{\cos \left(x+\frac{\pi }{4}\right)}\left[\because \cos (a+b)=\cos .\cos b-\sin a.\sin b\right]\\ =\frac{1}{\sqrt{2}}\int sec\left(x+\frac{\pi }{4}\right)dx\\ =\frac{1}{\sqrt{2}}\log \left|\tan \left(\frac{\pi }{4}+\frac{x}{2}+\frac{\pi }{8}\right)\right|+C\\ =\frac{1}{\sqrt{2}}\log \left|\tan \left(\frac{x}{2}+\frac{3\pi }{8}\right)\right|+C\end{array}\end{array}$

Hence, the correct answer is Option (D).