Question

$\int \sqrt{\left(1+\sin \left(\frac{x}{4}\right)\right)}dx=$

• A. $8\left[\sin \left(\frac{x}{8}\right)+\cos \left(\frac{x}{8}\right)\right]+c$
• B. $8\left[\sin \left(\frac{x}{8}\right)-\cos \left(\frac{x}{8}\right)\right]+c$
• C. $8\left[\sin \left(\frac{x}{8}\right)+\sin \left(\frac{x}{8}\right)\right]+c$
• D. $\left(\frac{1}{8}\right)\left[\sin \left(\frac{x}{8}\right)-\cos \left(\frac{x}{8}\right)\right]+c$

Answer 1

The correct option is B

$8\left[\sin \left(\frac{x}{8}\right)-\cos \left(\frac{x}{8}\right)\right]+c$

Explanation for the correct option:

Integrating the given function:

Consider the given equation as,

$$\begin{gathered} I=\int \sqrt{\left(1+\sin \left(\frac{x}{4}\right)\right)}dx \\ I=\int \sqrt{\left(1+\sin \left(2\left(\frac{x}{8}\right)\right)\right)}dx \end{gathered}$$

We know that,

$\sin 2x=2\sin x\cos x$

Then,

$I=\int \sqrt{\left(1+2\sin \left(\frac{x}{8}\right)\cos \left(\frac{x}{8}\right)\right)}dx$

We know that, ${\sin }^{2}x+{\cos }^{2}x=1$

Then,

$$\begin{gathered} I=\int \sqrt{\left({\sin }^2\left(\frac{x}{8}\right)+{\cos }^2\left(\frac{x}{8}\right)+2\sin \left(\frac{x}{8}\right)\cos \left(\frac{x}{8}\right)\right)}dx \\ =\int \sqrt{{\left(\sin \left(\frac{x}{8}\right)+\cos \left(\frac{x}{8}\right)\right)}^2}dx \\ =\int \sin \left(\frac{x}{8}\right)+\cos \left(\frac{x}{8}\right)dx \\ =\frac{-\cos \left({\displaystyle \frac{x}{8}}\right)}{{\displaystyle \frac{1}{8}}}+\frac{\sin \left({\displaystyle \frac{x}{8}}\right)}{{\displaystyle \frac{1}{8}}}+c\left[\because \int \sin xdx=-\cos xand\int \cos xdx=\sin x\right] \\ =-8\cos \left(\frac{x}{8}\right)+8\sin \left(\frac{x}{8}\right)+c \\ I=8\left[\sin \left(\frac{x}{8}\right)-\cos \left(\frac{x}{8}\right)\right]+c \end{gathered}$$

Hence, the correct answer is Option (B).