int_0^fracπ2cos^3x sqrtsinxdx

Explanation:

Type explanation here...

Another Explanation (5): সমাধান: ধরি, \( sinx = t^2 \) ∴ \( cosx dx = 2t dt \) \( \int_{0}^{\frac{\pi}{2}} cos^3x \sqrt{sinx} dx \) \( = \int_{0}^{\frac{\pi}{2}} cos^2x \sqrt{sinx} cosx dx \) \( = \int_{0}^{\frac{\pi}{2}} (1-sin^2x) \sqrt{sinx} cosx dx \) এখন, \( sinx = t^2 \) হলে, \( cosx dx = 2t dt \) যখন \( x = 0 \), \( t = 0 \) এবং যখন \( x = \frac{\pi}{2} \), \( t = 1 \) সুতরাং, \( \int_{0}^{\frac{\pi}{2}} (1-sin^2x) \sqrt{sinx} cosx dx \) \( = \int_{0}^{1} (1-t^4) \sqrt{t^2} 2t dt \) \( = 2 \int_{0}^{1} (1-t^4) t^2 dt \) \( = 2 \int_{0}^{1} (t^2 - t^6) dt \) \( = 2 \left[ \frac{t^3}{3} - \frac{t^7}{7} \right]_{0}^{1} \) \( = 2 \left[ \frac{1}{3} - \frac{1}{7} \right] \) \( = 2 \left[ \frac{7-3}{21} \right] \) \( = 2 \left[ \frac{4}{21} \right] \) \( = \frac{8}{21} \) 🎉