int_0^{pi / 4} sqrt{1-sin 2 x} ,d x = | vedclass

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

Question

(C) We know that $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$.So,$1 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\cos x - \sin x)

Vedclass · Accepted Answer

$\sqrt{2}-1$

Vedclass · Answer

(C) We know that $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$.So,$1 - \sin 2x = \sin^2 x + \cos^2 x - 2 \sin x \cos x = (\cos x - \sin x)^2$.The integral becomes $\int_0^{\pi / 4} \sqrt{(\cos x - \sin x)^2} \,d x = \int_0^{\pi / 4} |\cos x - \sin x| \,d x$.In the interval $[0, \pi / 4]$,$\cos x \geq \sin x$,so $|\cos x - \sin x| = \cos x - \sin x$.Thus,the integral is $\int_0^{\pi / 4} (\cos x - \sin x) \,d x = [\sin x + \cos x]_0^{\pi / 4}$.Evaluating the limits: $(\sin(\pi / 4) + \cos(\pi / 4)) - (\sin 0 + \cos 0) = (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1$.

$\int_0^{\pi / 4} \sqrt{1-\sin 2 x} \,d x =$

Similar Questions

Let $a$ and $b$ be real constants such that the function $f$ defined by $f(x) = \begin{cases} x^2+3x+a, & x \leq 1 \\ bx+2, & x > 1 \end{cases}$ is differentiable on $\mathbb{R}$. Then,the value of $\int_{-2}^2 f(x) dx$ equals

$\int_0^{2/3} \frac{dx}{4 + 9x^2} = $

Evaluate the following definite integral as a limit of sums:
$\int_{0}^{5}(x+1) d x$

$\int_0^1 x \operatorname{Sin}^{-1} x \, dx =$

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module