mathop smallint limits_0^pi sqrt {1 + 4{{sin | vedclass

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

Question

(B) The given integral is $I = \int_0^\pi \sqrt{1 + 4\sin^2\frac{x}{2} - 4\sin\frac{x}{2}} \; dx$.This can be written as $I = \int_0^\pi \sqrt{(1 - 2\

Vedclass · Accepted Answer

$4\sqrt 3 - 4 - \frac{\pi }{3}$

Vedclass · Answer

(B) The given integral is $I = \int_0^\pi \sqrt{1 + 4\sin^2\frac{x}{2} - 4\sin\frac{x}{2}} \; dx$.This can be written as $I = \int_0^\pi \sqrt{(1 - 2\sin\frac{x}{2})^2} \; dx = \int_0^\pi |1 - 2\sin\frac{x}{2}| \; dx$.We know $1 - 2\sin\frac{x}{2} = 0$ when $\sin\frac{x}{2} = \frac{1}{2}$,which means $\frac{x}{2} = \frac{\pi}{6}$,so $x = \frac{\pi}{3}$.For $0 \le x  0$. For $\frac{\pi}{3} Thus,$I = \int_0^{\pi/3} (1 - 2\sin\frac{x}{2}) \; dx + \int_{\pi/3}^\pi -(1 - 2\sin\frac{x}{2}) \; dx$.$I = [x + 4\cos\frac{x}{2}]_0^{\pi/3} - [x + 4\cos\frac{x}{2}]_{\pi/3}^\pi$.$I = ((\frac{\pi}{3} + 4\cos\frac{\pi}{6}) - (0 + 4\cos 0)) - ((\pi + 4\cos\frac{\pi}{2}) - (\frac{\pi}{3} + 4\cos\frac{\pi}{6}))$.$I = (\frac{\pi}{3} + 4(\frac{\sqrt{3}}{2}) - 4) - ((\pi + 0) - (\frac{\pi}{3} + 4(\frac{\sqrt{3}}{2})))$.$I = (\frac{\pi}{3} + 2\sqrt{3} - 4) - (\pi - \frac{\pi}{3} - 2\sqrt{3}) = \frac{\pi}{3} + 2\sqrt{3} - 4 - \frac{2\pi}{3} + 2\sqrt{3} = 4\sqrt{3} - 4 - \frac{\pi}{3}$.

$\mathop \smallint \limits_0^\pi \sqrt {1 + 4{{\sin }^2}\frac{x}{2} - 4\sin \frac{x}{2}} \;dx = $

Similar Questions

The value of the definite integral $\int\limits_{\infty}^{0} \frac{z e^{-z}}{\sqrt{1-e^{-2z}}} \, dz$ is:

$\int_{0}^{3} \sqrt{9 - x^2} dx =$

$\int_0^{\pi / 4} \frac{d x}{\cos ^3(x) \cdot \sqrt{2 \sin (2 x)}}=$

The value of the integral $\int_0^1 e^{x^2} dx$ lies in the interval

Choose the correct option regarding the following definite integrals:
$(i)$ $\int_0^{\pi / 2} \sin ^m(x) \cos (x) d x = \frac{1}{m+1}$
(ii) $\int_0^{\pi / 2} \sin (x) \cos ^n(x) d x = \frac{1}{n+1}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module