Question

The value of ${\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{2}}\frac{1+\sin \left(2x\right)+\cos \left(2x\right)}{\sin \left(x\right)+\cos \left(x\right)}dx=$

• A. $16$
• B. $8$
• C. $4$
• D. $1$

Answer 1

The correct option is D

$1$

Explanation for the correct option:

Compute the required value:

Given: ${\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{2}}\frac{1+\sin \left(2x\right)+\cos \left(2x\right)}{\sin \left(x\right)+\cos \left(x\right)}dx$

$$\begin{gathered} ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{2}}\frac{1+2\sin \left(x\right)\cos \left(x\right)+2{\cos }^2\left(x\right)-1}{\sin \left(x\right)+\cos \left(x\right)}dx\left[\sin \left(2x\right)=2\sin \left(x\right)\cos \left(x\right),\cos \left(2x\right)=2{\cos }^2\left(x\right)-1\right] \\ ={\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{2}}\frac{2\left(\cos \right(x\left)\right)\left[\sin \left(x\right)+\cos \left(x\right)\right]}{\sin \left(x\right)+\cos \left(x\right)}dx \\ =2{\int }_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{2}}\cos \left(x\right)·dx \\ =2{\left|\sin \left(x\right)\right|}_{\frac{\mathrm{\pi }}{6}}^{\frac{\mathrm{\pi }}{2}} \\ =2\left|\sin \left(\frac{\mathrm{\pi }}{2}\right)-\sin \left(\frac{\mathrm{\pi }}{6}\right)\right| \\ =2\times \left|1-\frac{1}{2}\right| \\ =2\times \frac{1}{2} \\ =1 \end{gathered}$$

Hence, option $D$ is the correct answer.