Question

The integral ${\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}}\frac{dx}{1+\cos x}$ is equal to

• A. $2$
• B. $4$
• C. $-1$
• D. $-2$

Answer 1

The correct option is A

$2$

Explanation for the correct option:

$$\begin{gathered} {\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}}\frac{dx}{2{\cos }^2\left({\displaystyle \frac{x}{2}}\right)}=\frac{1}{2}{\int }_{\frac{\mathrm{\pi }}{4}}^{\frac{3\mathrm{\pi }}{4}}se{c}^2\frac{x}{2}dx \\ =\frac{1}{2}{{\left[\frac{\tan \frac{x}{2}}{{\displaystyle \frac{1}{2}}}\right]}_{\frac{\mathrm{\pi }}{4}}}^{\frac{3\mathrm{\pi }}{4}} \\ =\tan \frac{3\mathrm{\pi }}{8}-\tan \frac{\mathrm{\pi }}{8} \\ \left[\tan \frac{\mathrm{\pi }}{8}=\sqrt{\frac{1-\cos \frac{\mathrm{\pi }}{4}}{1+\cos \frac{\mathrm{\pi }}{4}}}=\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}=\frac{\sqrt{2}-1}{1} \\ \tan \frac{3\mathrm{\pi }}{8}=\sqrt{\frac{1-\cos \frac{3\mathrm{\pi }}{4}}{1+\cos \frac{3\mathrm{\pi }}{4}}}=\sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}=\sqrt{2}+1\right] \\ =(\sqrt{2}+1)-(\sqrt{2}-1) \\ =2 \end{gathered}$$

Hence option (A) is correct.