int_(0)^(π/2)cos^3xsqrtsinxdx এর মান কত ?

প্রশ্ন: \( \int_{0}^{\pi/2} \cos^3x \sqrt{\sin x} \, dx \) এর মান কত?

সমাধান:

ধরি, \( I = \int_{0}^{\pi/2} \cos^3x \sqrt{\sin x} \, dx \)

আমরা \(\cos^2 x = 1 - \sin^2 x\) লিখতে পারি। সুতরাং,

\( I = \int_{0}^{\pi/2} \cos^2x \cdot \cos x \cdot \sqrt{\sin x} \, dx \)

\( I = \int_{0}^{\pi/2} (1 - \sin^2 x) \sqrt{\sin x} \cos x \, dx \)

ধরি, \( \sin x = t \). তাহলে, \( \cos x \, dx = dt \).

যখন \( x = 0 \), তখন \( t = \sin 0 = 0 \).

যখন \( x = \pi/2 \), তখন \( t = \sin (\pi/2) = 1 \).

সুতরাং, নতুন ইন্টিগ্রালটি হবে:

\( I = \int_{0}^{1} (1 - t^2) \sqrt{t} \, dt \)

\( I = \int_{0}^{1} (1 - t^2) t^{1/2} \, dt \)

\( I = \int_{0}^{1} (t^{1/2} - t^{5/2}) \, dt \)

এখন, ইন্টিগ্রেশন করি:

\( I = \left[ \frac{t^{3/2}}{3/2} - \frac{t^{7/2}}{7/2} \right]_{0}^{1} \)

\( I = \left[ \frac{2}{3} t^{3/2} - \frac{2}{7} t^{7/2} \right]_{0}^{1} \)

সীমা বসিয়ে পাই:

\( I = \left( \frac{2}{3} (1)^{3/2} - \frac{2}{7} (1)^{7/2} \right) - \left( \frac{2}{3} (0)^{3/2} - \frac{2}{7} (0)^{7/2} \right) \)

\( I = \frac{2}{3} - \frac{2}{7} \)

\( I = \frac{14 - 6}{21} \)

\( I = \frac{8}{21} \)

অতএব, \( \int_{0}^{\pi/2} \cos^3x \sqrt{\sin x} \, dx = \frac{8}{21} \). 🎉