intdx/(1+cos3x)=? 

প্রশ্ন: \(\int \frac{dx}{1+\cos 3x} = ?\)

সমাধান:

আমরা জানি, \(\cos 2\theta = 2\cos^2 \theta - 1\), সুতরাং, \(\cos 3x = 2\cos^2 \frac{3x}{2} - 1\).

অতএব, \(1 + \cos 3x = 1 + 2\cos^2 \frac{3x}{2} - 1 = 2\cos^2 \frac{3x}{2}\).

তাহলে, \[ \int \frac{dx}{1+\cos 3x} = \int \frac{dx}{2\cos^2 \frac{3x}{2}} = \frac{1}{2} \int \sec^2 \frac{3x}{2} dx \]

আমরা জানি, \(\int \sec^2 ax \, dx = \frac{1}{a} \tan ax + C\).

সুতরাং, \[ \frac{1}{2} \int \sec^2 \frac{3x}{2} dx = \frac{1}{2} \cdot \frac{1}{\frac{3}{2}} \tan \frac{3x}{2} + C = \frac{1}{2} \cdot \frac{2}{3} \tan \frac{3x}{2} + C = \frac{1}{3} \tan \frac{3x}{2} + C \]

সুতরাং, \(\int \frac{dx}{1+\cos 3x} = \frac{1}{3} \tan \frac{3x}{2} + C\).