y= $2^x \cdot cos^2\;x \cdot sin \; x \; \; Let\; u=2^x \; and \; v=cos^2\;x \cdot sin \; x \\ \; \\ \; \\ \; \\ u=2^x \\ \; \\ \; \\ \therefore diff. \; w.r.t. \; x, \; \; \therefore \; u_1=2^x(log 2)^1 \\ \; \\ \; \\ u_2 \;=\; 2^x (log2)(log2) \;=\; 2^x(log2)^2 \\ \; \\ \; \\ u_3 \; =\; 2^x (log2)^3 \\ \; \\ \; \\ Diff \; w.r.t. \; x \; 'n' \; times, \; \; \therefore u_n\; =\; 2^x (log2)^n \\ \; \\ \; \\ \; \\ v=cos^2\;x \cdot sin \; x \;= \dfrac{1}{2} (2cos^2\;x \cdot sin \; x) \;= \; \dfrac{1}{2} ( 2sin \; x cos\;x \cdot cos\;x) \\ \; \\ \; \\ \therefore v=\dfrac{1}{2}\{sin2xcosx\} \; =\; \dfrac{1}{4}\{2sin2xcosx\} \\ \; \\ \; \\ \therefore v=\dfrac{1}{4}\{sin3x+sinx\} \; \ldots {\because sin(A+B)+sin(A-B)\;=\;2sinAcosB} \\ \; \\ \; \\ v_1 = \dfrac{1}{4} \{ 3cos3x+cosx \} \; = \; \dfrac{1}{4} \bigg\{ 3sin(1.\frac{\pi}{2}-3x) + sin(1.\frac{\pi}{2}-x) \bigg\} \\ \; \\ \; \\ v_2 = \dfrac{1}{4} \{ -3^2sin3x-sinx \} \; = \; \dfrac{(-1)^3}{4} \bigg\{ 3^2sin(2.\frac{\pi}{2}-3x) + sin(2.\frac{\pi}{2}-x) \bigg\} \\ \; \\ \; \\ v_3 = \dfrac{1}{4} \{ -3^3cos3x-cosx \} \; = \; \dfrac{(-1)^4}{4} \bigg\{ 3^3sin(3.\frac{\pi}{2}-3x) + sin(3.\frac{\pi}{2}-x) \bigg\} \\ \; \\ \; \\ \therefore v_n \;=\; \dfrac{(-1)^{n+1}}{4} \bigg\{ 3^n sin(n.\frac{\pi}{2}-3x) + sin(n.\frac{\pi}{2}-x) \bigg\} \\ \; \\ \; \\ \; \\ Now, \; by \; Lebnitz’s \; theorem, \\ \; \\ \; \\ y_n \;=\; (uv)_n \;=\; ^nC_{0}u_{n}v + ^nC_{1}u_{n-1}v_{1}+ ^nC_{2}u_{n-2}v_{2} + ^nC_{3}u_{n-3}v_{3} + \ldots + ^nC_{n}uv_{n} \\ \; \\ \; \\ \; \\ \therefore y_n \;=\; 2^x(log2)^n \dfrac{1}{4}(sin3x+sinx) \; + \; \\ n.2^x.(log2)^{n-1} \dfrac{(-1)^2}{4} \bigg\{ 3sin(1.\frac{\pi}{2}-3x) + sin(1.\frac{\pi}{2}-x) \bigg\} \;+\; \\ \dfrac{n(n-1)}{2} 2^x.(log2)^{n-2} \dfrac{(-1)^3}{4} \bigg\{ 3^2 sin(2.\frac{\pi}{2}-3x) + sin(2.\frac{\pi}{2}-x) \bigg\} \\ + \ldots \ldots + 2^x.(log2) \dfrac{(-1)^{n+1}}{4} \bigg\{ 3^n sin(n.\frac{\pi}{2}-3x) + sin(n.\frac{\pi}{2}-x) \bigg\} $