Question:

<div class='solution'> <p><strong>Using Wallis formula:</strong> $\displaystyle\int_0^{\pi/2}\sin^m x\cos^n x\,dx = \frac{(m-1)\!\!(n-1)\!\!}{(m+n)\!\!}$ when $m+n$ is odd, no extra factor.</p> <p>Here $m=5$, $n=4$: $m+n=9$ (odd) → no $\pi/2$ factor.</p> <p>$(m-1)\!\! = 4\!\! = 4\cdot2 = 8$</p> <p>$(n-1)\!\! = 3\!\! = 3\cdot1 = 3$</p> <p>$(m+n)\!\! = 9\!\! = 9\cdot7\cdot5\cdot3\cdot1 = 945$</p> <p>$$I = \frac{8\cdot3}{945} = \frac{24}{945} = \frac{8}{315}$$</p> <p><strong>Verification by substitution:</strong> Let $t=\cos x$:</p> <p>$$I = \int_0^1(1-t^2)^2 t^4\,dt = \int_0^1(t^4-2t^6+t^8)\,dt = \frac{1}{5}-\frac{2}{7}+\frac{1}{9} = \frac{63-90+35}{315} = \frac{8}{315}\checkmark$$</p> </div>