Question

For $n\epsilon N,$ the value of the definite integral ${\int }_0^{n\pi +V}\sqrt{\frac{1+cos2x}{2}}dx$ where $\frac{\pi }{2}<V<\pi$ is

• A. $(2n+1)cosV$
• B. $2N-sinV$
• C. $2n+2-sinV$
• D. $2n+1-sinV$

Answer 1

The correct option is A $(2n+1)cosV$

Let $I={\int }_0^{n\pi +V}\sqrt{\frac{1+\cos 2x}{2}}dx$

$={\int }_0^{n\pi +V}\sqrt{\frac{2{\cos }^{2}x}{2}}dx$

$={\int }_0^{n\pi +V}\cos xdx$

$={\left[\sin x\right]}_0^{n\pi }$

$=\sin \left(n\pi +v\right)-\sin 0$

$=\sin \left(n\pi +v\right)$

$=\left(2n+1\right).\cos V$

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