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

If $g(1) = g(2)$,then the value of $\int_{1}^{2} [f\{g(x)\}]^{-1} f'\{g(x)\} g'(x) dx$ is:

The value of the integral $\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$ is equal to

$\int\limits_0^3 \left( \frac{1}{\sqrt{x^2 + 4x + 4}} + \sqrt{x^2 - 4x + 4} \right) dx =$

If $\int_0^{\frac{\pi}{3}} \cos^4 x \, dx = a\pi + b\sqrt{3}$,where $a$ and $b$ are rational numbers,then $9a + 8b$ is equal to:

Evaluate the definite integral: $\int_0^{\pi /2} \cos^2 x \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module