int(sinx+cosx)/(sqrt(1+sin2x))dx=? 

প্রশ্ন: \(\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx = ?\)

সমাধান:

আমরা জানি, \(\sin 2x = 2 \sin x \cos x\)

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

অতএব, \(\sqrt{1 + \sin 2x} = \sqrt{(\sin x + \cos x)^2} = |\sin x + \cos x|\)

সুতরাং, \(\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx = \int \frac{\sin x + \cos x}{|\sin x + \cos x|} dx\)

যদি \(\sin x + \cos x > 0\) হয়, তবে:

\(\int \frac{\sin x + \cos x}{\sin x + \cos x} dx = \int 1 dx = x + c\)

যদি \(\sin x + \cos x < 0\) হয়, তবে:

\(\int \frac{\sin x + \cos x}{-(\sin x + \cos x)} dx = \int -1 dx = -x + c\)

সাধারণভাবে, আমরা লিখতে পারি:

\(\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx = \int \frac{\sin x + \cos x}{|\sin x + \cos x|} dx = x + c\), যখন \(\sin x + \cos x > 0\)

\(\int \frac{\sin x + \cos x}{\sqrt{1 + \sin 2x}} dx = \int \frac{\sin x + \cos x}{|\sin x + \cos x|} dx = -x + c\), যখন \(\sin x + \cos x < 0\)

উত্তর: \(x + c\)

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