Question

If $x\epsilon (\pi ,\frac{3\pi }{2})$ then $\int (\sqrt{1+sinx}-\sqrt{1-sinx})dx=$

• A. $4cos\frac{x}{2}+c$
• B. $-4cos\frac{x}{2}+c$
• C. $4sin\frac{x}{2}+c$
• D. $-4sin\frac{x}{2}+c$

Answer 1

The correct option is A $4cos\frac{x}{2}+c$

Let $=\int \sqrt{{\left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)}^2}-\sqrt{{\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}^2}dx$

$=\int \left(\cos \frac{x}{2}+\sin \frac{x}{2}\right)-\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)dx$

$=\int \cos \frac{x}{2}+\sin \frac{x}{2}-\cos \frac{x}{2}+\sin \frac{x}{2}dx$

$=\int 2\sin \frac{x}{2}dx$

$=\frac{-2\times \cos \frac{x}{2}}{\frac{1}{2}}+c$

$=-4\cos \frac{x}{2}+c$

Option $\left(B\right)$ is correct answer.