int_0^{32 pi} sqrt{1-cos 4 x} , dx = (in sqrt | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that $1 - \cos 4x = 2 \sin^2(2x)$.Therefore, $\sqrt{1 - \cos 4x} = \sqrt{2 \sin^2(2x)} = \sqrt{2} |\sin 2x|$.The integral becomes $\int_0^

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(D) We know that $1 - \cos 4x = 2 \sin^2(2x)$.Therefore, $\sqrt{1 - \cos 4x} = \sqrt{2 \sin^2(2x)} = \sqrt{2} |\sin 2x|$.The integral becomes $\int_0^{32 \pi} \sqrt{2} |\sin 2x| \, dx = \sqrt{2} \int_0^{32 \pi} |\sin 2x| \, dx$.Since $|\sin 2x|$ is a periodic function with period $\frac{\pi}{2}$, and the interval $[0, 32 \pi]$ contains $\frac{32 \pi}{\pi/2} = 64$ periods.Thus, $\int_0^{32 \pi} |\sin 2x| \, dx = 64 \int_0^{\pi/2} \sin 2x \, dx$.Evaluating the integral: $64 \left[ -\frac{\cos 2x}{2} \right]_0^{\pi/2} = 64 \left( -\frac{\cos \pi}{2} - (-\frac{\cos 0}{2}) \right) = 64 \left( \frac{1}{2} + \frac{1}{2} \right) = 64$.Multiplying by the constant $\sqrt{2}$, the final result is $64 \sqrt{2}$.

$\int_0^{32 \pi} \sqrt{1-\cos 4 x} \, dx =$ (in $\sqrt{2}$)

Similar Questions

$\int_{-\pi}^{\frac{\pi}{2}} \sin x \cdot \sin^2(\cos x) \, dx =$

Let $f(x) = \max \{x+|x|, x-[x]\}$, where $[x]$ denotes the greatest integer $\leq x$. Then, the value of $\int_{-3}^{3} f(x) dx$ is

If $\phi(t)=\begin{cases} 1, & \text{for } 0 \leq t < 1 \\ 0, & \text{otherwise} \end{cases}$, then $\int_{-3000}^{3000} \left( \sum_{r'=2014}^{2016} \phi(t-r') \phi(t-2016) \right) dt$ is

If $\int_3^5 \sqrt{8 x-x^2-15} d x=p$,then $\sin p+\operatorname{cosec} p=$

$\int\limits_2^4 {\left[ {{{\log }_x}2 - \frac{{{{\left( {{{\log }_x}2} \right)}^2}}}{{\ln 2}}} \right]} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module