int_(-pi/2)^(pi/2)(sinx+cosx)^2dx=?

প্রশ্ন: \(\int_{-\pi/2}^{\pi/2} (\sin x + \cos x)^2 dx = ?\)

সমাধান:

আমরা জানি, \((a+b)^2 = a^2 + 2ab + b^2\)

সুতরাং, \((\sin x + \cos x)^2 = \sin^2 x + 2 \sin x \cos x + \cos^2 x\)

আমরা আরও জানি, \(\sin^2 x + \cos^2 x = 1\) এবং \(2 \sin x \cos x = \sin 2x\)

সুতরাং, \((\sin x + \cos x)^2 = 1 + \sin 2x\)

এখন, \(\int_{-\pi/2}^{\pi/2} (\sin x + \cos x)^2 dx = \int_{-\pi/2}^{\pi/2} (1 + \sin 2x) dx\)

\(= \int_{-\pi/2}^{\pi/2} 1 dx + \int_{-\pi/2}^{\pi/2} \sin 2x dx\)

\(= [x]_{-\pi/2}^{\pi/2} + \left[-\frac{\cos 2x}{2}\right]_{-\pi/2}^{\pi/2}\)

\(= \left[\frac{\pi}{2} - \left(-\frac{\pi}{2}\right)\right] + \left[-\frac{\cos (2 \cdot \frac{\pi}{2})}{2} - \left(-\frac{\cos (2 \cdot (-\frac{\pi}{2}))}{2}\right)\right]\)

\(= \left[\frac{\pi}{2} + \frac{\pi}{2}\right] + \left[-\frac{\cos \pi}{2} + \frac{\cos (-\pi)}{2}\right]\)

\(= \pi + \left[-\frac{-1}{2} + \frac{-1}{2}\right]\)

\(= \pi + \left[\frac{1}{2} - \frac{1}{2}\right]\)

\(= \pi + 0\)

\(= \pi\)

অতএব, \(\int_{-\pi/2}^{\pi/2} (\sin x + \cos x)^2 dx = \pi\)