নিচের যোগজের মান নির্ণয় করঃint_(π/3)^(π/2) cos^5x/sin^7x dx 

Explanation:

Another Explanation (5): যোগজের মান নির্ণয়: \(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\cos^5 x}{\sin^7 x} dx\) ধরি, \(I = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\cos^5 x}{\sin^7 x} dx\) \(I = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\cos^5 x}{\sin^5 x \cdot \sin^2 x} dx\) \(I = \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \cot^5 x \csc^2 x dx\) এখন, ধরি \(\cot x = z\) তাহলে, \(-\csc^2 x dx = dz\) সুতরাং, \(\csc^2 x dx = -dz\) যখন \(x = \frac{\pi}{3}\), তখন \(z = \cot \frac{\pi}{3} = \frac{1}{\sqrt{3}}\) যখন \(x = \frac{\pi}{2}\), তখন \(z = \cot \frac{\pi}{2} = 0\) তাহলে, \(I = \int_{\frac{1}{\sqrt{3}}}^{0} z^5 (-dz)\) \(I = -\int_{\frac{1}{\sqrt{3}}}^{0} z^5 dz\) \(I = \int_{0}^{\frac{1}{\sqrt{3}}} z^5 dz\) \(I = \left[ \frac{z^6}{6} \right]_{0}^{\frac{1}{\sqrt{3}}}\) \(I = \frac{1}{6} \left[ \left(\frac{1}{\sqrt{3}}\right)^6 - 0^6 \right]\) \(I = \frac{1}{6} \left[ \frac{1}{(\sqrt{3})^6} \right]\) \(I = \frac{1}{6} \left[ \frac{1}{3^3} \right]\) \(I = \frac{1}{6} \cdot \frac{1}{27}\) \(I = \frac{1}{162}\) অতএব, \(\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\cos^5 x}{\sin^7 x} dx = \frac{1}{162}\) 🎉