Q.Evaluate the following:

Step 1. (i) $\displaystyle\int_0^{\pi/2}\sin^{10}x\,dx$ — $n=10$ even. Closed form: $\dfrac{9\cdot7\cdot5\cdot3\cdot1}{10\cdot8\cdot6\cdot4\cdot2}\cdot\dfrac{\pi}{2}$. Numerator $=945$, denominator $=3840$, so $\dfrac{945}{3840}=\dfrac{63}{256}$ (dividing by $15$). Result: $\dfrac{63}{256}\cdot\dfrac{\pi}{2}=\dfrac{63\pi}{512}$.

Step 2. (ii) $\displaystyle\int_0^{\pi/2}\cos^7x\,dx$ — $n=7$ odd. Closed form (no $\pi/2$ factor for an odd index): $\dfrac{6\cdot4\cdot2}{7\cdot5\cdot3\cdot1}=\dfrac{48}{105}=\dfrac{16}{35}$.

Step 3. (iii) $\displaystyle\int_0^{\pi/4}\sin^62x\,dx$ — substitute first. Let $u=2x,\ du=2dx$; limits $x{:}0\to\pi/4\Rightarrow u{:}0\to\pi/2$. So $\displaystyle\int_0^{\pi/4}\sin^62x\,dx=\dfrac12\int_0^{\pi/2}\sin^6u\,du$. With $n=6$ even: $\dfrac{5\cdot3\cdot1}{6\cdot4\cdot2}\cdot\dfrac{\pi}{2}=\dfrac{15}{48}\cdot\dfrac{\pi}{2}=\dfrac{5}{16}\cdot\dfrac{\pi}{2}=\dfrac{5\pi}{32}$. Multiplying by the $\tfrac12$ from the substitution: $\dfrac12\cdot\dfrac{5\pi}{32}=\dfrac{5\pi}{64}$.

Step 4. (iv) $\displaystyle\int_0^{\pi/6}\sin^53x\,dx$ — substitute first. Let $u=3x,\ du=3dx$; limits $x{:}0\to\pi/6\Rightarrow u{:}0\to\pi/2$. So $\displaystyle\int_0^{\pi/6}\sin^53x\,dx=\dfrac13\int_0^{\pi/2}\sin^5u\,du$. With $n=5$ odd: $\dfrac{4\cdot2}{5\cdot3\cdot1}=\dfrac{8}{15}$. Multiplying by $\tfrac13$: $\dfrac13\cdot\dfrac{8}{15}=\dfrac{8}{45}$.

Step 5. (v) $\displaystyle\int_0^{\pi/2}\sin^2x\cos^4x\,dx$ — $m=2,n=4$, both even. Closed form includes the $\pi/2$ factor since both are even: $\dfrac{(m-1)(m-3)\cdots\,\cdot\,(n-1)(n-3)\cdots}{(m+n)(m+n-2)\cdots}\cdot\dfrac{\pi}{2}$. For $m=2$: the descending odd product is just $1$. For $n=4$: $3\cdot1=3$. Numerator $=1\cdot3=3$. Denominator, $m+n=6$: $6\cdot4\cdot2=48$. So $\dfrac{3}{48}\cdot\dfrac{\pi}{2}=\dfrac1{16}\cdot\dfrac{\pi}2=\dfrac{\pi}{32}$. (Cross-check via the Beta function: $\tfrac12B(\tfrac32,\tfrac52)=\tfrac12\cdot\dfrac{\Gamma(3/2)\Gamma(5/2)}{\Gamma(4)}=\tfrac12\cdot\dfrac{(\sqrt\pi/2)(3\sqrt\pi/4)}{6}=\tfrac12\cdot\dfrac{3\pi/8}{6}=\dfrac{\pi}{32}$ ✓.)

Step 6. (vi) $\displaystyle\int_0^{2\pi}\sin^7\dfrac{x}{4}\,dx$ — substitute first. Let $u=x/4,\ du=dx/4\Rightarrow dx=4\,du$; limits $x{:}0\to2\pi\Rightarrow u{:}0\to\pi/2$. So $\displaystyle\int_0^{2\pi}\sin^7\dfrac{x}{4}dx=4\int_0^{\pi/2}\sin^7u\,du$. With $n=7$ odd (same coefficient as Step 2): $\dfrac{16}{35}$. Multiplying by $4$: $4\cdot\dfrac{16}{35}=\dfrac{64}{35}$. …

Reduction-formula / Wallis closed forms for $\int_0^{\pi/2}\sin^nx\,dx$, $\cos^nx\,dx$, $\sin^mx\cos^nx\,dx$, and $\int_0^1x^m(1-x …

• Forgetting the extra scaling factor ($\tfrac12,\tfrac13$ or $4$) introduced by the substitutions $u=2x,3x,x/4$ in parts (iii), (iv), (vi)
• Applying the 'both even $\Rightarrow\times\pi/2$' rule when one of $m,n$ is actually odd — confusing part (v) (both even, has $\pi/2$) with part (vii) (both odd, purely rational) …