Question:

<b>Method - 1 :</b><br>Put $\cos x = t$; $-\sin x dx = dt$.<br>so that $I = \int \sin^3 x \cos^5 x dx = -\int (1-t^2) t^5 dt$<br>$= \int (t^7 - t^5) dt = \frac{t^8}{8} - \frac{t^6}{6} = \frac{\cos^8 x}{8} - \frac{\cos^6 x}{6} + C$<br><br><b>Method - 2 :</b><br>Put $\sin x = t$; $\cos x dx = dt$<br>so that $I = \int t^3 (1-t^2)^2 dt = \int (t^3 - 2t^5 + t^7) dt$<br>$= \frac{\sin^4 x}{4} - \frac{2\sin^6 x}{6} + \frac{\sin^8 x}{8} + C$<br><br><b>Method - 3 :</b><br>$\sin^3 x \cos^5 x = \frac{1}{8} \sin^3 2x \cos^2 x$<br>$= \frac{1}{64} (4 \sin^3 2x) (2 \cos^2 x)$<br>$= \frac{1}{64} (3 \sin 2x - \sin 6x) (1 + \cos 2x)$<br>$= \frac{1}{128} [6 \sin 2x + 6 \sin 2x \cos 2x - 2 \sin 6x - 2 \sin 6x \cos 2x]$<br>$= \frac{1}{128} [6 \sin 2x + 5 \sin 4x - 2 \sin 6x - \sin 8x]$<br>$\int \sin^3 x \cos^5 x dx = \frac{1}{128} \left[ \frac{\cos 6x}{3} + \frac{\cos 8x}{8} - \frac{5 \cos 4x}{4} - 3 \cos 2x \right] + C$