int_0^(pi/2)cos2xcos3xdx এর মান কত ?

প্রশ্ন: \( \int_0^{\frac{\pi}{2}} \cos 2x \cos 3x \, dx \) এর মান কত?

আমরা জানি, \( 2 \cos A \cos B = \cos(A+B) + \cos(A-B) \)।

সুতরাং, \( \cos 2x \cos 3x = \frac{1}{2} [\cos(2x+3x) + \cos(2x-3x)] \)

\( = \frac{1}{2} [\cos 5x + \cos(-x)] \)

\( = \frac{1}{2} [\cos 5x + \cos x] \) (\(\because \cos(-x) = \cos x\))

এখন, ইন্টিগ্রালটি হবে:

\( \int_0^{\frac{\pi}{2}} \cos 2x \cos 3x \, dx = \int_0^{\frac{\pi}{2}} \frac{1}{2} (\cos 5x + \cos x) \, dx \)

\( = \frac{1}{2} \int_0^{\frac{\pi}{2}} (\cos 5x + \cos x) \, dx \)

\( = \frac{1}{2} \left[ \int_0^{\frac{\pi}{2}} \cos 5x \, dx + \int_0^{\frac{\pi}{2}} \cos x \, dx \right] \)

আমরা জানি, \( \int \cos ax \, dx = \frac{\sin ax}{a} + C \)।

সুতরাং, \( \int_0^{\frac{\pi}{2}} \cos 5x \, dx = \left[ \frac{\sin 5x}{5} \right]_0^{\frac{\pi}{2}} = \frac{\sin \frac{5\pi}{2}}{5} - \frac{\sin 0}{5} = \frac{\sin \left(2\pi + \frac{\pi}{2}\right)}{5} - 0 = \frac{\sin \frac{\pi}{2}}{5} = \frac{1}{5} \)

এবং, \( \int_0^{\frac{\pi}{2}} \cos x \, dx = \left[ \sin x \right]_0^{\frac{\pi}{2}} = \sin \frac{\pi}{2} - \sin 0 = 1 - 0 = 1 \)

তাহলে, \( \int_0^{\frac{\pi}{2}} \cos 2x \cos 3x \, dx = \frac{1}{2} \left[ \frac{1}{5} + 1 \right] = \frac{1}{2} \left[ \frac{1+5}{5} \right] = \frac{1}{2} \cdot \frac{6}{5} = \frac{3}{5} \)

অতএব, \( \int_0^{\frac{\pi}{2}} \cos 2x \cos 3x \, dx = \frac{3}{5} \) 🥳