int_{0}^{pi/4} sqrt{1+sin 2x} dx = rule{1cm}{ | vedclass

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

Question

(B) We know that $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$.Therefore, $\sqrt{1+\sin 2x} = \sqrt{\sin^2 x + \cos^2 x + 2 \sin x \cos x}

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) We know that $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$.Therefore, $\sqrt{1+\sin 2x} = \sqrt{\sin^2 x + \cos^2 x + 2 \sin x \cos x} = \sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|$.Since $x \in [0, \pi/4]$, both $\sin x$ and $\cos x$ are non-negative, so $|\sin x + \cos x| = \sin x + \cos x$.Now, the integral becomes $\int_{0}^{\pi/4} (\sin x + \cos x) dx$.Integrating term by term, we get $[-\cos x + \sin x]_0^{\pi/4}$.Evaluating at the limits: $(-\cos(\pi/4) + \sin(\pi/4)) - (-\cos(0) + \sin(0))$.$= (-1/\sqrt{2} + 1/\sqrt{2}) - (-1 + 0) = 0 - (-1) = 1$.

$\int_{0}^{\pi/4} \sqrt{1+\sin 2x} dx = \rule{1cm}{0.15mm}$

Similar Questions

$ \int_{0}^{\frac{\pi}{2}} \frac{1}{a^{2} \sin ^{2} x+b^{2} \cos ^{2} x} d x $ is equal to

$\left[ \int_{0}^{2} \sqrt{x + \sqrt{x + \sqrt{x + \dots \infty}}} \, dx \right]$ is equal to (where $[\cdot]$ is the $G.I.F.$)

$\int_0^{2\pi} (\sin x + \cos x) \, dx = $

If $\int_{n}^{n+1} g(x) dx = n^2, \forall n \in Z$,then the value of $\int_{-3}^3 g(x) dx$ is

All values of $a$ for which the inequality $\frac{1}{\sqrt{a}} \int_{1}^{a} (\frac{3}{2} \sqrt{x} + 1 - \frac{1}{\sqrt{x}}) dx < 4$ is satisfied, lie in the interval

Vedclass Test Series

Exam Paper Generator

Online Exam Module